Rd Sharma 2021 for Class 10 Maths Chapter 5 - Arithmetic Progressions

Answer:

In the given problem, we need to write the first term (a) and the common difference (d) of the given A.P

(i) −5, −1, 3, 7 …

Here, first term of the given A.P is (a) = −5

Now, we will find the difference between the two terms of the given A.P

Similarly,

Also,

As

Therefore, the first term of the given A.P is and the common difference of the given is

(ii)

Here, first term of the given A.P is (a) =

Now, we will find the difference between the two terms of the given A.P

Similarly,

Also,

As

Therefore, the first term of the given A.P is and the common difference is

(iii) 0.3, 0.55, 0.80, 1.05, …

Here, first term of the given A.P is (a) = 0.3

Now, we will find the difference between the two terms of the given A.P

Similarly,

Also,

As

Therefore, the first term of A.P is and the common difference is

(iv) −1.1, −3.1, −5.1, −7.1...

Here, first term of the given A.P is (a) = −1.1

Now, we will find the difference between the two terms of the given A.P

Similarly,

Also,

As

Therefore, the first term of A.P is and the common difference is

Answer:

In the given problem, we are given its first term (a) and common difference (d).

We need to find the A.P

(i) ,

Now, as

A.P would be represented by

So,

Similarly,

Also,

Further,

Therefore, A.P with and is

(ii) ,

Now, as

A.P would be represented by

So,

Similarly,

Also,

Further,

Therefore, A.P with and is

(iii) ,

Now, as

A.P would be represented by

So,

Similarly,

Also,

Further,

Therefore, A.P with and is

Answer:

(i) In the given problem,

Cost of digging a well for the first meter = Rs 150

Cost of digging a well for subsequent meter is increased by Rs 20

So,
Cost of digging a well of depth one meter= Rs. 150

Cost of digging a well of depth two meters= Rs = Rs.

Cost of digging a well of depth three meters= Rs = Rs.

Cost of digging a well of depth four meters = Rs = Rs.

Thus, the costs of digging a well of different depths are

Now, for a sequence to be an A.P., the difference between adjacent terms should be equal.

Here,

Also,

Therefore,

Since the terms of the sequence are at a common difference of 20, the above sequence is an A.P. with the first term asand common difference.

(ii) Here, let us take the initial amount of air present in the cylinder as 100 units.

So,

Amount left after vacuum pump removes air for 1^st time=

Amount left after vacuum pump removes air for 2^nd time=

Amount left after vacuum pump removes air for 3^rd time=

Thus, the amount left in the cylinder at various stages is

Now, for a sequence to be an A.P., the difference between adjacent terms should be equal.

Here,

Also,

Since,

The sequence is not an A.P.

(iii)
Here, prinical (P) = 1000
Rate (r) = 10%
Amount compounded annually is given by
$A=P{\left(1+\frac{r}{100}\right)}^n$
For the first year,
$A_1=1000{\left(1+\frac{10}{100}\right)}^1=1100$
For the second year,
$A_2=1000{\left(1+\frac{10}{100}\right)}^2=1210$
For the third year,
$A_1=1000{\left(1+\frac{10}{100}\right)}^3=1331$
Therefore, first three terms are 1100, 1210, 1331.
The common difference between the consecutive terms are not same.
Hence, this is not in A.P.
 

Answer:

In the given problem, we need to find the common difference and the next four terms of the given A.P.

(i)

Here, first term (a₁) =1

Common difference (d)

Now, we need to find the next four terms of the given A.P

That is we need to find

So, using the formula

Substituting in the above formula

Substituting n = 5, we get

Substituting n = 6, we get

Substituting n = 7, we get

Substituting n = 8, we get

Therefore, the common difference is and the next four terms are

(ii)

Here, first term (a₁) =0

Common difference (d)

Now, we need to find the next four terms of the given A.P

That is we need to find

So, using the formula

Substituting in the above formula

Substituting n = 5, we get

Substituting n = 6, we get

Substituting n = 7, we get

Substituting n = 8, we get

Therefore, the common difference is and the next four terms are

(iii)

Here, first term (a₁) =−1

Common difference (d)

Now, we need to find the next four terms of the given A.P

That is we need to find

So, using the formula

Substituting in the above formula

Substituting n = 4, we get

Substituting n = 5, we get

Substituting n = 6, we get

Substituting n = 7, we get

Therefore, the common difference is and the next four terms are

(iv)

Here, first term (a₁) =−1

Common difference (d)

Now, we need to find the next four terms of the given A.P

That is we need to find

So, using the formula

Substituting in the above formula

Substituting n = 4, we get

$$\begin{gathered} a_4=-1+\left(4-1\right)\left(\frac{1}{6}\right) \\ {a}_4=-1+\left(\frac{1}{2}\right) \\ {a}_4=\frac{-2+1}{2}=\frac{-1}{2} \end{gathered}$$

Substituting n = 5, we get

Substituting n = 6, we get

Substituting n = 7, we get

Therefore, the common difference is and the next four terms are

Answer:

In the given problem, we are given various sequences.

We need to find out that the given sequences are an A.P or not and then find its common difference (d)

(i) 

Here,

First term (a) = 3

Now, for the given to sequence to be an A.P,

Common difference (d) 

Here,

Also,

Since 

Hence, the given sequence is not an A.P

(ii) 

Here,

First term (a) = 0

Now, for the given sequence to be an A.P,

Common difference (d) 

Here

Also,

Since 

Hence, the given sequence is an A.P and its common difference is 

(iii) 

Here,

First term (a) = 

Now, for the given to sequence to be an A.P,

Common difference (d) 

Here,

Also,

Since 

Hence, the given sequence is not an A.P

(iv) 

Here,

First term (a) = 12

Now, for the given to sequence to be an A.P,

Common difference (d) 

Here,

Also,

Since 

Hence, the given sequence is an A.P with the common difference 

(v) 

Here,

First term (a) = 3

Now, for the given to sequence to be an A.P,

Common difference (d) 

Here,

Also,

Since 

Hence, the given sequence is an A.P and its common difference is 

(vi) Where,

Here,

First term (a) = p

Now, for the given to sequence to be an A.P,

Common difference (d) 

Here,

Also,

Since 

Hence, the given sequence is an A.P and its common difference is 

(vii) 

Here,

First term (a) = 1.0

Now, for the given to sequence to be an A.P,

Common difference (d) 

Here,

Also,

Since 

Hence, the given sequence is an A.P and its common difference is 

(viii) 

Here,

First term (a) = −225

Now, for the given to sequence to be an A.P,

Common difference (d) 

Here,

Also,

Since 

Hence, the given sequence is an A.P and its common difference is 

(ix) 

Here,

First term (a) = 10

Now, for the given to sequence to be an A.P,

Common difference (d) = $a_1-a=a_2-a_1$ 

Here,

Also,

Since 

Hence, the given sequence is not an A.P

(x) 

Here,

First term (a) = a + b

Now, for the given to sequence to be an A.P,

Common difference (d) 

Here,

Also,

Since 

Hence, the given sequence is an A.P and its common difference is 

(xi) 

Here,

First term (a) = 

Now, for the given to sequence to be an A.P,

Common difference (d) 

Here,

Also,

Since 

Hence, the given sequence is not an A.P

(xii) 

Here,

First term (a) = 

Now, for the given to sequence to be an A.P,

Common difference (d) 

Here,

Also,

$a_3-a_2=7^2-5^2=49 - 25 = 24$

Since 

Hence, the given sequence is an A.P with the common difference .

Answer:

In this problem, we are given different A.P. and we need to find the common difference of the A.P., along with the next two terms.

(i)

Here,

So, common difference of the A.P. (d) =

Also, we need to find the next two terms of A.P., which means we have to find the 5^th and 6^th term.

So, for fifth term,

Similarly, we find the sixth term,

Therefore, the common difference is and the next two terms of the A.P. are.

(ii)

Here,

So, common difference of the A.P. (d) =

Also, we need to find the next two terms of A.P., which means we have to find the 5^th and 6^th term.

So, for fifth term,

Similarly, we find the sixth term,

Therefore, the common difference is and the next two terms of the A.P. are.

(iii)

Here,

So, common difference of the A.P. (d) =

Also, we need to find the next two terms of A.P., which means we have to find the 5^th and 6^th term.

So, for fifth term,

Similarly, we find the sixth term,

Therefore, the common difference is and the next two terms of the A.P. are.

(iv)

Here,

So, common difference of the A.P. (d) =

Also, we need to find the next two terms of A.P., which means we have to find the 5^th and 6^th term.

So, for fifth term,

Similarly, we find the sixth term,

Therefore, the common difference is and the next two terms of the A.P. are.

(v)

Here,

So, common difference of the A.P. (d) =

Also, we need to find the next two terms of A.P., which means we have to find the 5^th and 6^th term.

So, for fifth term,

Similarly, we find the sixth term,

Therefore, the common difference is and the next two terms of the A.P. are.

Answer:

In the given problem, we are given the sequence with the n^th term () as where a and b are real numbers.

We need to show that this sequence is an A.P and then find its common difference (d)

Here,

Now, to show that it is an A.P, we will find its few terms by substituting

So,

Substituting n = 1, we get

Substituting n = 2, we get

Substituting n = 3, we get

Further, for the given to sequence to be an A.P,

Common difference (d)

Here,

Also,

Since

Hence, the given sequence is an A.P and its common difference is .

Answer:

In this problem, we are given different A.P. and we need to find the required term of that A.P.

(i) 10^th term of the A.P.

Here,

First term (a) = 1

Common difference of the A.P. (d)

Now, as we know,

So, for 10^th term,

Therefore, the 10^th term of the given A.P. is.

(ii) 18^th term of the A.P.

Here,

First term (a) =

Common difference of the A.P. (d)

Now, as we know,

So, for 18^th term,

Therefore, the 18^th term of the given A.P. is.

(iii) n^th term of the A.P.

Here,

First term (a) = 13

Common difference of the A.P. (d)

Now, as we know,

So, for n^th term,

Therefore, the n^th term of the given A.P. is.

(iv) 10^th term of the A.P.

Here,

First term (a) = -40

Common difference of the A.P. (d)

Now, as we know,

So, for 10^th term,

Therefore, the 10^th term of the given A.P. is.

(v) 8^th term of the A.P.

Here,

First term (a) = 117

Common difference of the A.P. (d)

Now, as we know,

So, for 8^th term,

Therefore, the 8^th term of the given A.P. is.

(vi) 11^th term of the A.P.

Here,

First term (a) = 10.0

Common difference of the A.P. (d)

Now, as we know,

So, for 11^th term,

Therefore, the 11^th term of the given A.P. is.

(vii) 9^th term of the A.P.

Here,

First term (a) =

Common difference of the A.P. (d)

Now, as we know,

So, for 9^th term,

Therefore, the 9^th term of the given A.P. is.

Answer:

In the given problem, we are given an A.P and the value of one of its term.

We need to find which term it is (n)

So here we will find the value of n using the formula, 

(i) Here, A.P is 

Now,

Common difference (d) = 

=

= 5

Thus, using the above mentioned formula

Thus, 

Therefore 248 is the  of the given A.P

(ii) Here, A.P is 

Now,

Common difference (d) = 

=80-84

= -4

Thus, using the above mentioned formula

On further simplifying, we get,

Thus, 

Therefore 84 is the  of the given A.P

(iii) Here, A.P is 

Now,

Common difference (d) = 

= 9-4

= 5

Thus, using the above mentioned formula

Thus, 

Therefore 254 is the  of the given A.P

(iv) Here, A.P is 

Now,

Common difference (d) = 

= 42-21

= 21

Thus, using the above mentioned formula

Thus, 

Therefore 420 is the  of the given A.P

(v) Here, A.P is 

We need to find first negative term of the A.P

Now,

Common difference (d) = 

Now, we need to find the first negative term,

Further simplifying, we get,

Thus, 

Therefore, the first negative term is the of the given A.P.

Answer:

In the given problem, we are given an A.P and the value of one of its term.

We need to find whether it is a term of the A.P or not.

So here we will use the formula,

(i) Here, A.P is

Now,

Common difference (d) =

Thus, using the above mentioned formula, we get,

Since, the value of n is a fraction.

Thus, 68 is not the term of the given A.P

Therefore the answer is

(ii) Here, A.P is

Now,

Common difference (d) =

Thus, using the above mentioned formula, we get,

Since, the value of n is a fraction.

Thus, 302 is not the term of the given A.P

Therefore the answer is

(iii) Here, A.P is

Now,

Common difference (d) =

Thus, using the above mentioned formula

Since, the value of n is a fraction.

Thus, -150 is not the term of the given A.P

Therefore, the answer is

Answer:

In the given problem, we are given an A.P.

We need to find the number of terms present in it

So here we will find the value of n using the formula,

(i) Here, A.P is

The first term (a) = 7

The last term () = 43

Now,

Common difference (d) =
                                      

Thus, using the above mentioned formula, we get,

Thus,

Therefore, the number of terms present in the given A.P is .

(ii) Here, A.P is

The first term (a) = -1

The last term () =

Now,

Common difference (d) =
                                      

Thus, using the above mentioned formula, we get,

Further solving for n, we get

Thus,

Therefore, the number of terms present in the given A.P is .

(iii) Here, A.P is

The first term (a) = 7

The last term () = 205

Now,

Common difference (d) =
                                     

Thus, using the above mentioned formula, we get,

Thus,

Therefore, the number of terms present in the given A.P is

(iv) Here, A.P is

The first term (a) = 18

The last term () = -47

Now,

Common difference (d) =
                                      

Thus, using the above mentioned formula, we get,

Further, solving for n, we get

Thus,

Therefore, the number of terms present in the given A.P is .

Answer:

In the given problem, we are given an A.P whose,

First term (a) = 5

Last term () = 80

Common difference (d) = 3

We need to find the number of terms present in it (n)

So here we will find the value of n using the formula,

So, substituting the values in the above mentioned formula

Thus,

Therefore, the number of terms present in the given A.P is

Answer:

In the given problem, we are given 6^th and 17^th term of an A.P.

We need to find the 40^th term

Here,

Now, we will find and using the formula

So,

Also,

So, to solve for a and d

On subtracting (1) from (2), we get

Substituting (3) in (1), we get

Thus,

Substituting the above values in the formula

Therefore,

Answer:

In the given problem, we are given 10^th and 18^th term of an A.P.

We need to find the 26^th term

Here,

Now, we will find and using the formula

So,

Also,

So, to solve for a and d

On subtracting (1) from (2), we get

Substituting d=4 in (1), we get

Thus,

Substituting the above values in the formula,

Therefore,

Answer:

Here, we are given two A.P. sequences whose n^th terms are equal. We need to find n. 

So let us first find the n^th term for both the A.P.

First A.P. is 9, 7, 5 …

Here,

First term (a) = 9

Common difference of the A.P. (d)

Now, as we know,

So, for n^th term,

Second A.P. is 15, 12, 9 …

Here,

First term (a) = 15

Common difference of the A.P. (d)

Now, as we know,

So, for n^th term,

Now, we are given that the n^th terms for both the A.P. sequences are equal, we equate (1) and (2),

Therefore,

Answer:

In the given problem, we need to find the 12^th term from the end for the given A.P.

(i) 3, 5, 7, 9 …201

Here, to find the 12^th term from the end let us first find the total number of terms. Let us take the total number of terms as n.

So,

First term (a) = 3

Last term (a_n) = 201

Common difference (d) =

Now, as we know,

So, for the last term,

Further simplifying,

So, the 12^th term from the end means the 89^th term from the beginning.

So, for the 89^th term (n = 89)

Therefore, the 12^th term from the end of the given A.P. is.

(ii) 3, 8, 13 …253

Here, to find the 12^th term from the end let us first find the total number of terms. Let us take the total number of terms as n.

So,

First term (a) = 3

Last term (a_n) = 253

Common difference, d =

Now, as we know,

So, for the last term,

Further simplifying,

So, the 12^th term from the end means the 40^th term from the beginning.

So, for the 40^th term (n = 40)

Therefore, the 12^th term from the end of the given A.P. is.

(iii) 1, 4, 7, 10 …88

Here, to find the 12^th term from the end let us first find the total number of terms. Let us take the total number of terms as n.

So,

First term (a) = 1

Last term (a_n) = 88

Common difference, d= =3

Now, as we know,

So, for the last term,

Further simplifying,

So, the 12^th term from the end means the 19^th term from the beginning.

So, for the 19^th term (n = 19)

Therefore, the 12^th term from the end of the given A.P. is.

Answer:

In the given problem, the 9^th term of an A.P. is zero.

Here, let us take the first term of the A.P as a and the common difference as d

So, as we know,

We get,

Now, we need to prove that 29^th term is double of 19^th term. So, let us first find the two terms.

For 19^th term (n = 19),

(Using 1)

For 29^th term (n = 29),

$$\begin{gathered} a_{29}=a+\left(29-1\right)d \\ =-8d+28d \\ =20d \\ =2\times 10d \\ =2\times a_{19} \end{gathered}$$  (Using 1)

Therefore, for the given A.P. the 29^th term is double of the 19^th term.

Hence proved.

Answer:

Here, let us take the first term of the A.P. as a and the common difference as d

We are given that 10 times the 10^th term is equal to 15 times the 15^th term. We need to show that 25^th term is zero.

So, let us first find the two terms.

So, as we know,

For 10^th term (n = 10),

For 15^th term (n = 15),

Now, we are given,

Solving this, we get,

Next, we need to prove that the 25^th term of the A.P. is zero. For that, let us find the 25^th term using n = 25,

Thus, the 25^th term of the given A.P. is zero.

Hence proved

Answer:

Here, we are given that 24^th term is twice the 10^th term, for a certain A.P. Here, let us take the first term of the A.P. as a and the common difference as d

We have to prove that

So, let us first find the two terms.

As we know,

For 10^th term (n = 10),

For 24^th term (n = 24),

Now, we are given that

So, we get,

Further, we need to prove that the 72^nd term is twice of 34^th term. So let now find these two terms,

For 34^th term (n = 34),

(Using 1)

For 72^nd term (n = 72),

(Using 1)

Therefore,

Answer:

It is given that, $a_{26}=0, a_{11}=3, a_n=-\frac{1}{5}$.
$$\begin{gathered} \Rightarrow a+25d=0 .....\left(i\right) \\ \Rightarrow a+10d=3 .....\left(ii\right) \\ \mathrm{Subtracting} \left(ii\right) \mathrm{from} \left(i\right). \\ 15d=-3 \\ \Rightarrow d=-\frac{1}{5} \\ \Rightarrow a=-25d=5 \\ Now, a_n=a+\left(n-1\right)d \\ \Rightarrow -\frac{1}{5}=5+\left(n-1\right)\left(-\frac{1}{5}\right) \\ \Rightarrow -\frac{1}{5}-5=\left(n-1\right)\left(-\frac{1}{5}\right) \\ \Rightarrow n=27 \end{gathered}$$

Answer:

In the given problem, let us take the first term as a and the common difference as d

Here, we are given that,

We need to find a and d

So, as we know,

For the 4^th term (n = 4),

Similarly, for the 3^rd term (n = 3),

Also, for the 7^th term (n = 7),

Now, using the value of a₃ in equation (2), we get,

Equating (3) and (4), we get,

On further simplification, we get,

Now, to find a,

Therefore, for the given A.P

Answer:

In the given problem, we are given 6^th and 8^th term of an A.P.

We need to find the 2^nd and n^th term

Here, let us take the first term as a and the common difference as d

We are given,

Now, we will find and using the formula

So,

Also,

So, to solve for a and d

On subtracting (1) from (2), we get

Substituting (3) in (1), we get

Thus,

So, for the 2^nd term (n = 2),

For the n^th term,

Therefore,

Answer:

In this problem, we need to find out how many numbers of two digits are divisible by 3.

So, we know that the first two digit number that is divisible by 3 is 12 and the last two digit number divisible by 3 is 99. Also, all the terms which are divisible by 3 will form an A.P. with the common difference of 3.

So here,

First term (a) = 12

Last term (a_n) = 99

Common difference (d) = 3

So, let us take the number of terms as n

Now, as we know,

So, for the last term,

Further simplifying,

Therefore, the number of two digit terms divisible by 3 is.

Answer:

In the given problem, we need to find the 32^nd term of an A.P. which contains a total of 60 terms.

Here we are given the following,

First term (a) = 7

Last term (a_n) = 125

Number of terms (n) = 60

So, let us take the common difference as d

Now, as we know,

So, for the last term,

Further simplifying,

So, for the 32^nd term (n = 32)

Therefore, the 32^nd term of the given A.P. is.

Answer:

In the given problem, the sum of 4^th and 8^th term is 24 and the sum of 6^th and 10^th term is 34.

We can write this as,

We need to find a and d

For the given A.P., let us take the first term as a and the common difference as d

As we know,

For 4^th term (n = 4),

For 8^th term (n = 8),

So, on substituting the above values in (1), we get,

Also, for 6^th term (n = 6),

For 10^th term (n = 10),

So, on substituting the above values in (2), we get,

Next we simplify (3) and (4). On subtracting (3) from (4), we get,

Further, using the value of d in equation (3), we get,

On further simplifying, we get,

Therefore, for the given A.P

Answer:

In the given problem, we are given 1^st and 100^th term of an A.P.

We need to find the 50^th term

Here,

Now, we will find d using the formula

So,

Also,

So, to solve for d

Substituting a = 5, we get

Thus,

Substituting the above values in the formula,

Therefore,

Answer:

In this problem, we are given different A.P. and we need to find.

(i) A.P.

Here,

First term (a) = -9

Common difference of the A.P. (d)

Now, as we know,

Here, we find and

So, for 30^th term,

Also, for 20^th term,

So,

Therefore, for the given A.P

(ii) A.P.

Here,

First term (a) = a

Common difference of the A.P. (d) =d

Now, as we know,

Here, we find a₃₀ and a₂₀.

So, for 30^th term,

Also, for 20^th term,

So,

Therefore, for the given A.P

Answer:

A.P: a, a+d, a+2d …

Here, we first need to write the expression for

Now, as we know,

So, for n^th term,

Similarly, for k^th term

So,

So,

(i) In the given problem, we are given 11^th and 13^th term of an A.P.

We need to find the common difference. Let us take the common difference as d and the first term as a.

Here,

Now, we will find and using the formula

So,

Also,

Solving for a and d

On subtracting (1) from (2), we get

Therefore, the common difference for the A.P. is.

(ii) We are given,

Here,

Let us take the first term as a and the common difference as d

Now, as we know,

Here, we find a₃₀ and a₂₀.

So, for 10^th term,

Also, for 5^th term,

So,

Therefore, the common difference for the A.P. is.

(iii) In the given problem, the 20^th term is 10 more than the 18^th term. So, let us first find the 20^th term and 18^th term of the A.P.

Here

Let us take the first term as a and the common difference as d

Now, as we know,

So, for 20^th term (n = 20),

Also, for 18^th term (n = 18),

Now, we are given,

On substituting the values, we get,

Therefore, the common difference for the A.P. is

Answer:

In the given problem, we need to find the number of terms in an A.P

(i) 25, 50, 75, 100 …

We are given,

Let us take the total number of terms as n.

So,

First term (a) = 25

Last term (a_n) = 1000

Common difference (d) =

Now, as we know,

So, for the last term,

Therefore, the total number of terms of the given A.P. is.

(ii) -1, -3, -5, -7 …

We are given,

Let us take the total number of terms as n.

So,

First term (a) = −1

Last term (a_n) = −151

Common difference (d) =

Now, as we know,

So, for the last term,

On further simplifying, we get,

Therefore, the total number of terms of the given A.P. is.

(iii)

We are given,

Let us take the total number of terms as n.

So,

First term (a) =

Last term (a_n) = 550

Common difference (d) =

Now, as we know,

So, for the last term,

On further simplifying, we get,

Therefore, the total number of terms of the given A.P. is

_(iv)

We are given,

Let us take the total number of terms as n.

So,

First term (a) = 1

Last term (a_n) =

Common difference (d) =

Now, as we know,

So, for the last term,

On further simplifying, we get,

Therefore, the total number of terms of the given A.P. is.

Answer:

$$\begin{gathered} a_8=\frac{1}{2}{a}_2 and a_{11}=\frac{1}{3}{a}_4+1 \\ \Rightarrow a+7d=\frac{\left(a+d \right)}{2} \\ \Rightarrow a+13d=0 .....\left(i\right) \end{gathered}$$
$$\begin{gathered} And, \\ a_{11}=\frac{1}{3}{a}_4+1 \\ \Rightarrow a+10d=\frac{a+3d}{3}+1 \\ \Rightarrow 2a+27d=1 .....\left(ii\right) \\ Solve \left(i\right) and \left(ii\right), we get d = 1. \\ Therefore, a = -13 \\ {a}_{15}=a+14d=-13+14=1 \end{gathered}$$

Answer:

Here, let us take the first term of the A.P as a and the common difference of the A.P as d

Now, as we know,

So, for 3^rd term (n = 3),

Also, for 5^th term (n = 5),

For 7^th term (n = 7),

Now, we are given,

Substituting the value of d in (1), we get,

So, the first term is 4 and the common difference is 6.

Therefore, the A.P. is

Answer:

Here, let us take the first term of the A.P. as a and the common difference of the A.P as d

Now, as we know,

So, for 7^th term (n = 7),

Also, for 13^th term (n = 13),

Now, on subtracting (2) from (1), we get,

Substituting the value of d in (1), we get,

So, the first term is 2 and the common difference is 5.

Therefore, the A.P. is

Answer:

In the given problem, let us first find the 13^th term of the given A.P.

A.P. is 3, 10, 17 …

Here,

First term (a) = 3

Common difference of the A.P. (d) =7

Now, as we know,

So, for 13^th term (n = 13),

Let us take the term which is 84 more than the 13^th term as a_n. So,

Also,

Further simplifying, we get,

Therefore, theof the given A.P. is 84 more than the 13^th term.

Answer:

Here, we are given two A.P sequences which have the same common difference. Let us take the first term of one A.P. as a and of other A.P. as a’

Also, it is given that the difference between their 100^th terms is 100.

We need to find the difference between their 100^th terms

So, let us first find the 100^th terms for both of them.

Now, as we know,

So, for 100^th term of first A.P. (n = 100),

Now, for 100^th term of second A.P. (n = 100),

Now, we are given,

On substituting the values, we get,

Now, we need the difference between the 1000^th terms of both the A.P.s

So, for 1000^th term of first A.P. (n = 1000),

Now, for 1000^th term of second A.P. (n = 1000),

So,

Therefore, the difference between the 1000^th terms of both the arithmetic progressions will be.

Answer:

Here, we are given two A.P. sequences. We need to find the value of n for which the n^th terms of both the sequences are equal. We need to find n

So let us first find the n^th term for both the A.P.

First A.P. is 63, 65, 67 …

Here,

First term (a) = 63

Common difference of the A.P. (d) =2

Now, as we know,

So, for n^th term,

Second A.P. is 3, 10, 17 …

Here,

First term (a) = 3

Common difference of the A.P. (d) =7

Now, as we know,

So, for n^th term,

Now, we are given that the n^th terms for both the A.P. sequences are equal, we equate (1) and (2),

Therefore,

Answer:

In this problem, we need to find out how many multiples of 4 lie between 10 and 250.

So, we know that the first multiple of 4 after 10 is 12 and the last multiple of 4 before 250 is 248. Also, all the terms which are divisible by 4 will form an A.P. with the common difference of 4.

So here,

First term (a) = 12

Last term (a_n) = 248

Common difference (d) = 4

So, let us take the number of terms as n

Now, as we know,

So, for the last term,

Further simplifying,

Therefore, the number of multiples of 4 that lie between 10 and 250 is.

Answer:

In this problem, we need to find out how many numbers of three digits are divisible by 7.

So, we know that the first three digit number that is divisible by 7 is 105 and the last three digit number divisible by 7 is 994. Also, all the terms which are divisible by 7 will form an A.P. with the common difference of 7.

So here,

First term (a) = 105

Last term (a_n) = 994

Common difference (d) = 7

So, let us take the number of terms as n

Now, as we know,

So, for the last term,

Further simplifying,

Therefore, the number of three digit terms divisible by 7 is.

Answer:

In the given problem, let us first find the 41^st term of the given A.P.

A.P. is 8, 14, 20, 26 …

Here,

First term (a) = 8

Common difference of the A.P. (d) =6

Now, as we know,

So, for 41^st term (n = 41),

Let us take the term which is 72 more than the 41^st term as a_n. So,

Also,

Further simplifying, we get,

Therefore, the of the given A.P. is 72 more than the 41^st term.

Answer:

In the given problem, let us first find the 36^st term of the given A.P.

A.P. is 9, 12, 15, 18 …

Here,

First term (a) = 9

Common difference of the A.P. (d) =3

Now, as we know,

So, for 36^th term (n = 36),

Let us take the term which is 39 more than the 36^th term as a_n. So,

Also,

Further simplifying, we get,

Therefore, theof the given A.P. is 39 more than the 36^th term

Answer:

In the given problem, we need to find the 8^th term from the end for the given A.P.

We have the A.P as 7, 10, 13 …184

Here, to find the 8^th term from the end let us first find the total number of terms. Let us take the total number of terms as n.

So,

First term (a) = 7

Last term (a_n) = 184

Common difference (d) = =3

Now, as we know,

So, for the last term,

Further simplifying,

So, the 8^th term from the end means the 53^rd term from the beginning.

So, for the 53^rd term (n = 53)

Therefore, the 8^th term from the end of the given A.P. is.

Answer:

In the given problem, we need to find the 10^th term from the end for the given A.P.

We have the A.P as 8, 10, 12 …126

Here, to find the 10^th term from the end let us first find the total number of terms. Let us take the total number of terms as n.

So,

First term (a) = 8

Last term (a_n) = 126

Common difference (d) = =2

Now, as we know,

So, for the last term,

Further simplifying,

So, the 10^th term from the end means the 51^st term from the beginning.

So, for the 51^st term (n = 51)

Therefore, the 10^th term from the end of the given A.P. is.

Answer:

In the given problem, the sum of 4^th and 8^th term is 24 and the sum of 6^th and 10^th term is 44. We have to find the A.P

We can write this as,

We need to find the A.P

For the given A.P., let us take the first term as a and the common difference as d

As we know,

For 4^th term (n = 4),

For 8^th term (n = 8),

So, on substituting the above values in (1), we get,

Also, for 6^th term (n = 6),

For 10^th term (n = 10),

So, on substituting the above values in (2), we get,

Next we simplify (3) and (4). On subtracting (3) from (4), we get,

Further, using the value of d in equation (3), we get,

So here,

Therefore, the A.P. is.

Answer:

In the given problem, let us first find the 21^st term of the given A.P.

A.P. is 3, 15, 27, 39 …

Here,

First term (a) = 3

Common difference of the A.P. (d) =12

Now, as we know,

So, for 21^st term (n = 21),

Let us take the term which is 120 more than the 21^st term as a_n. So,

Also,

Further simplifying, we get,

Therefore, theof the given A.P. is 120 more than the 21^st term.

Answer:

Let a be the first term and d be the common difference.

We know that, n^th term = a_n = a + (n − 1)d

According to the question,

a₁₇ = 5 + 2a₈
⇒ a + (17 − 1)d =  5 + 2(a + (8 − 1)d)
⇒ a + 16d =  5 + 2a + 14d
⇒ 16d − 14d =  5 + 2a − a
⇒ 2d =  5 + a
⇒ a = 2d − 5    .... (1)

Also, a₁₁ = 43
⇒ a + (11 − 1)d = 43
⇒ a + 10d = 43   ....(2)

On substituting the values of (1) in (2), we get
2d − 5 + 10d = 43
⇒ 12d = 5 + 43
⇒ 12d = 48
⇒ d = 4
⇒ a = 2 × 4 − 5     [From (1)]
⇒ a = 3

∴ a_n = a + (n − 1)d
        = 3 + (n − 1)4
        = 3 + 4n − 4
        = 4n − 1

Thus, the n^th term of the given A.P. is  4n − 1.

Answer:

First three-digit number that is divisible by 9 is 108.

Next number is 108 + 9 = 117.

And the last three-digit number that is divisible by 9 is 999.

Thus, the progression will be 108, 117, .... , 999.

All are three digit numbers which are divisible by 9, and thus forms an A.P. having first term a 108 and the common difference as 9.

We know that, n^th term = a_n = a + (n − 1)d

According to the question,

999 = 108 + (n − 1)9
⇒ 108 + 9n − 9 = 999
⇒ 99 + 9n = 999
⇒ 9n = 999 − 99
⇒ 9n = 900
⇒ n = 100

Thus, the number of all three digit natural numbers which are divisible by 9 is 100.

Answer:

Let a be the first term and d be the common difference.

We know that, n^th term = a_n = a + (n − 1)d

According to the question,

a₁₉ = 3a₆
⇒ a + (19 − 1)d =  3(a + (6 − 1)d)
⇒ a + 18d =  3a + 15d
⇒ 18d − 15d =  3a − a
⇒ 3d = 2a
⇒ a = $\frac{3}{2}$d   .... (1)

Also, a₉ = 19
⇒ a + (9 − 1)d = 19
⇒ a + 8d = 19   ....(2)

On substituting the values of (1) in (2), we get
$\frac{3}{2}$d + 8d = 19
⇒ 3d + 16d = 19 × 2
⇒ 19d = 38
⇒ d = 2
⇒ a = $\frac{3}{2}\times 2$     [From (1)]
⇒ a = 3

Thus, the A.P. is 3, 5, 7, 9, .... .

Answer:

Let a be the first term and d be the common difference.

We know that, n^th term = a_n = a + (n − 1)d

According to the question,

a₉ = 6a₂
⇒ a + (9 − 1)d =  6(a + (2 − 1)d)
⇒ a + 8d =  6a + 6d
⇒ 8d − 6d =  6a − a
⇒ 2d = 5a
⇒ a = $\frac{2}{5}$d   .... (1)

Also, a₅ = 22
⇒ a + (5 − 1)d = 22
⇒ a + 4d = 22   ....(2)

On substituting the values of (1) in (2), we get
$\frac{2}{5}$d + 4d = 22
⇒ 2d + 20d = 22 × 5
⇒ 22d = 110
⇒ d = 5
⇒ a = $\frac{2}{5}\times 5$     [From (1)]
⇒ a = 2

Thus, the A.P. is 2, 7, 12, 17, .... .

Answer:

Let a be the first term and d be the common difference.

We know that, n^th term = a_n = a + (n − 1)d

According to the question,

a₂₄ = 2a₁₀
⇒ a + (24 − 1)d = 2(a + (10 − 1)d)
⇒ a + 23d = 2a + 18d
⇒ 23d − 18d = 2a − a
⇒ 5d = a
⇒ a = 5d   .... (1)

Also,
a₇₂ = a + (72 − 1)d
      = 5d + 71d              [From (1)]
      = 76d                      ..... (2)

and
a₁₅ = a + (15 − 1)d
      = 5d + 14d              [From (1)]
      = 19d                      ..... (3)

On comparing (2) and (3), we get

76d = 4 × 19d
⇒ a₇₂ = 4 × a₁₅

Thus, 72^nd term of the given A.P. is 4 times its 15^th term.

Answer:

Since, the number is divisible by both 2 and 5, means it must be divisible by 10.

In the given numbers, first number that is divisible by 10 is 110.

Next number is 110 + 10 = 120.

The last number that is divisible by 10 is 990.

Thus, the progression will be 110, 120, ..., 990.

 All the terms are divisible by 10, and thus forms an A.P. having first term as 110 and the common difference as 10.

We know that, n^th term = a_n = a + (n − 1)d

According to the question,

990 = 110 + (n − 1)10
⇒ 990 = 110 + 10n − 10
⇒ 10n = 990 − 100
⇒ 10n = 890
⇒ n = 89

Thus, ​the number of natural numbers between 101 and 999 which are divisible by both 2 and 5 is 89.

Answer:

Let a be the first term and d be the common difference.

We know that, n^th term = a_n = a + (n − 1)d

According to the question,

a₇ = $\frac{1}{9}$
⇒ a + (7 − 1)d = $\frac{1}{9}$
⇒ a + 6d = $\frac{1}{9}$       .... (1)

Also, a₉ = $\frac{1}{7}$
⇒ a + (9 − 1)d = $\frac{1}{7}$
⇒ a + 8d = $\frac{1}{7}$   ....(2)

On Subtracting (1) from (2), we get
8d − 6d = $\frac{1}{7}-\frac{1}{9}$
⇒ 2d = $\frac{9-7}{63}$
⇒ 2d = $\frac{2}{63}$
⇒ d = $\frac{1}{63}$
⇒ a = $\frac{1}{9}-\frac{6}{63}$     [From (1)]
⇒ a = $\frac{7-6}{63}$
⇒ a = $\frac{1}{63}$

∴ a₆₃ = a + (63 − 1)d
        = $\frac{1}{63}+\frac{62}{63}$
        = $\frac{63}{63}$
        = 1

Thus, (63)^rd term of the given A.P. is 1.

Answer:

Let a be the first term and d be the common difference.

We know that, n^th term = a_n = a + (n − 1)d

According to the question,

a₅ + a₉ = 30
⇒ a + (5 − 1)d + a + (9 − 1)d = 30
⇒ a + 4d + a + 8d = 30
⇒ 2a + 12d = 30
⇒ a + 6d = 15       .... (1)

Also, a₂₅ = 3(a₈)
⇒ a + (25 − 1)d = 3[a + (8 − 1)d]
⇒ a + 24d = 3a + 21d
⇒ 3a − a = 24d − 21d
⇒ 2a = 3d
⇒ a = $\frac{3}{2}$d   ....(2)

Substituting the value of (2) in (1), we get
$\frac{3}{2}$d + 6d = 15
⇒ 3d + 12d = 15 × 2
⇒ 15d = 30
⇒ d = 2
⇒ a = $\frac{3}{2}\times 2$         [From (1)]
⇒ a = 3

Thus, the A.P. is 3, 5, 7, 9, .... .

Answer:

Let a be the first term and d be the common difference.

We know that, n^th term = a_n = a + (n − 1)d

It is given that a = 40, d = −3 and a_n = 0

According to the question,

⇒ 0 = 40 + (n − 1)(−3)
⇒ 0 = 40 − 3n + 3
⇒ 3n = 43
⇒ n = $\frac{43}{3}$          .... (1)

Here, n is the number of terms, so must be an integer.

Thus, there is no term where 0 (zero) is a term of the A.P. 40, 37, 34, 31, ..... .

Answer:

Let a be the first term and d be the common difference.

We know that, n^th term = a_n = a + (n − 1)d

It is given that a = 213, d = −8 and a_n = 37

According to the question,

⇒ 37 = 213 + (n − 1)(−8)
⇒ 37 = 213 − 8n + 8
⇒ 8n = 221 − 37
⇒ 8n = 184
⇒ n = 23          .... (1)

Therefore, total number of terms is 23.

Since, there are odd number of terms.
So, Middle term will be ${\left(\frac{23+1}{2}\right)}^{\mathrm{th}}$ term, i.e., the 12^th term.

∴ a₁₂ = 213 + (12 − 1)(−8)
          = 213 − 88 
         = 125

Thus, the middle term of the A.P. 213, 205, 197, ...., 37 is 125.

Answer:

Let a be the first term and d be the common difference.

We know that, n^th term = a_n = a + (n − 1)d

According to the question,

a₅ = 31
⇒ a + (5 − 1)d =  31
⇒ a + 4d =  31
⇒ a = 31 − 4d        .... (1)

Also, a₂₅ = 140 + a₅
⇒ a + (25 − 1)d = 140 + 31
⇒ a + 24d = 171    .... (2)

On substituting the values of (1) in (2), we get
31 − 4d + 24d = 171
⇒ 20d = 171 − 31
⇒ 20d = 140
⇒ d = 7
⇒ a = 31 − 4 × 7     [From (1)]
⇒ a = 3

Thus, the A.P. is 3, 10, 17, 24, .... .

Answer:

$$\begin{gathered} a=-\frac{4}{3}, d=\frac{1}{3}, a_n=\frac{13}{3} \\ \Rightarrow a+\left(n-1\right)d=\frac{13}{3} \\ \Rightarrow \left(-\frac{4}{3}\right)+\left(n-1\right)\left(\frac{1}{3}\right)=\frac{13}{3} \\ \Rightarrow 13=-4+n-1 \\ \Rightarrow n=18 \\ \mathrm{Midlle} \mathrm{terms} \mathrm{are} {\frac{\mathrm{n}}{2}}^{\mathrm{th}} \mathrm{and} \frac{\mathrm{n}}{2}+1^{\mathrm{th}} , \mathrm{i}.\mathrm{e} 9\mathrm{th} \mathrm{and} 10\mathrm{th} \mathrm{terms}. \\ {a}_9=a+8d=-\frac{4}{3}+\frac{8}{3}=\frac{4}{3} \\ {a}_{10}=a+9d=-\frac{4}{3}+\frac{9}{3}=\frac{5}{3} \\ \therefore a_9+a_{10}=\frac{9}{3}=3 \end{gathered}$$

Answer:

Here, we are given that (m+1)^th term is twice the (n+1)^th term, for a certain A.P. Here, let us take the first term of the A.P. as a and the common difference as d

We need to prove that

So, let us first find the two terms.

As we know,

For (m+1)^th term (n’ = m+1)

For (n+1)^th term (n’ = n+1),

Now, we are given that

So, we get,

Further, we need to prove that the (3m+1)^th term is twice of (m+n+1)^th term. So let us now find these two terms,

For (m+n+1)^th term (n’ = m+n+1 ),

(Using 1)

For (3m+1)^th term (n’ = 3m+1),

Therefore,

Hence proved

Answer:

In the given problem, we have an A.P. which consists of n terms.

Here,

The first term (a) = a

The last term (a_n) = l

Now, as we know,

So, for the m^th term from the beginning, we take (n = m),

Similarly, for the m^th term from the end, we can take l as the first term.

So, we get,

Now, we need to prove

So, adding (1) and (2), we get,

Therefore,

Hence proved

Answer:

The given A.P is 11, 15, 19, .....,299.
$$\begin{gathered} a=11 \\ d=4 \\ {a}_n=299 \\ \Rightarrow a+\left(n-1\right)d=299 \\ \Rightarrow 11+\left(n-1\right)4=299 \\ \Rightarrow \left(n-1\right)4=288 \\ \Rightarrow n=73 \end{gathered}$$

Answer:

Consider the A.P −2, −4, −6, ...., −100.
$$\begin{gathered} a=-2, d=-2, a_n=-100 \\ \Rightarrow a+\left(n-1\right)d=-100 \\ \Rightarrow \left(-2\right)+\left(n-1\right)\left(-2\right)=-100 \\ \Rightarrow 2n=100 \\ \Rightarrow n=50 \\ \mathrm{The} 12\mathrm{th} \mathrm{term} \mathrm{from} \mathrm{the} \mathrm{end} \mathrm{will} \mathrm{be} \mathrm{the} 39\mathrm{th} \mathrm{term} \mathrm{from} \mathrm{the} \mathrm{starting}. \\ \therefore a_{39}=a+38d=-2+38\left(-2\right)=-78 \end{gathered}$$

Answer:

Consider the A.P −3, −7, −11, ....
$$\begin{gathered} a=-3, d=-4, \\ {a}_{30}-a_{20}=\left[a+\left(30-1\right)d\right]-\left[a+\left(20-1\right)d\right] \\ {a}_{30}-a_{20}=a+29d-a-19d \\ \therefore a_{30}-a_{20}=10d \\ =10\left(-4\right) \\ =-40 \end{gathered}$$

Answer:

First term of 1st A.P is 2.
First term of 2nd A.P is 7.
Consider the difference of their 10th terms,
$$\begin{gathered} a_{10}-a{\text{'}}_{10}=a+9d-a\text{'}-9d\text{'} \\ =a-a\text{'}+9d-9d\text{'} \\ =2-7+0 \left[d=d\text{'}\right] \\ =-5 \\ {a}_{21}-a{\text{'}}_{21}=a+20d-a\text{'}-20d\text{'} \\ =a-a\text{'}+20d-20d\text{'} \\ =2-7+0 \left[d=d\text{'}\right] \\ =-5 \\ Therefore, a_{10}-a{\text{'}}_{10}=a_{21}-a{\text{'}}_{21}. \end{gathered}$$
The difference between any two corresponding terms of A.P's is same as the difference between their terms.

Answer:

Here, we are given three terms,

First term (a₁) =

Second term (a₂) =

Third term (a₃) =

We need to find the value of x for which these terms are in A.P. So, in an A.P. the difference of two adjacent terms is always constant. So, we get,

…… (1)

Also,

…… (2)

Now, on equating (1) and (2), we get,

Therefore, for, these three terms will form an A.P.

Answer:

Here, we are given three terms which are in A.P.,

First term (a₁) =

Second term (a₂) =

Third term (a₃) =

We need to find the value of x. So, in an A.P. the difference of two adjacent terms is always constant. So, we get,

…… (1)

Also,

…… (2)

Now, on equating (1) and (2), we get,

Therefore, for, these three terms will form an A.P.

Answer:

Here, we are given three terms and we need to show that they are in A.P.,

First term (a₁) =

Second term (a₂) =

Third term (a₃) =

So, in an A.P. the difference of two adjacent terms is always constant. So, to prove that these terms are in A.P. we find the common difference, we get,

…… (1)

Also,

…… (2)

Now, since in equations (1) and (2) the values of d are equal, we can say that these terms are in A.P. with 2ab as the common difference.

Hence proved

Answer:

In the given problem, the sum of three terms of an A.P is 27 and the product of the three terms is 648. We need to find the three terms.

Here,

Let the three terms be where a is the first term and d is the common difference of the A.P

So,

…… (1)

Also,

Further solving for d,

…… (2)

Now, substituting (1) and (2) in three terms

First term =

So,

Also,

Second term = a

So,

Also,

Third term =

So,

Therefore, the three terms are .

Answer:

In the given problem, the sum of three terms of an A.P is 21 and the product of the first and the third term exceeds the second term by 6.

We need to find the three terms.

Here,

Let the three terms be where, a is the first term and d is the common difference of the A.P

So,

…… (1)

Also,

(Using)

(Using 1)

Further solving for d,

…… (2)

Now, using the values of a and d in the expressions of the three terms, we get,

First term =

So,

Second term = a

So,

Also,

Third term =

So,

Therefore, the three terms are .

Answer:

Here, we are given that four number are in A.P., such that there sum is 50 and the greatest number is 4 times the smallest.

So, let us take the four terms as.

Now, we are given that sum of these numbers is 50, so we get,

…… (1)

Also, the greatest number is 4 times the smallest, so we get,

…… (2)

Now, using (2) in (1), we get,

Now, using the value of a in (2), we get

So, first term is given by,

Second term is given by,

Third term is given by,

Fourth term is given by,

Therefore, the four terms are .

Answer:

In the given problem, the sum of three terms of an A.P is 12 and the sum of their cubes is 288.

We need to find the three terms.

Here,

Let the three terms be where, a is the first term and d is the common difference of the A.P

So,

Also, it is given that

So, using the properties:

We get,

Further solving for d by substituting the value of a, we get,

On further simplification, we get,

Now, here d can have two values +2 and -2.

So, on substituting the values of a = 4 and d = 2 in three terms, we get,

First term =

So,

Second term = a

So,

Third term =

So,

Also, on substituting the values of a = 4 and in three terms, we get,

First term =

So,

Second term = a

So,

Third term =

So,

Therefore, the three terms are .

Answer:

Let the four terms of the AP be a − 3d, a − d, a + d and a + 3d.
Given:
(a − 3d) + (a − d) + (a + d) + (a + 3d) = 56
$\Rightarrow$ 4a = 56
$\Rightarrow$ a = 14

$$\begin{gathered} \mathrm{Also}, \\ \frac{\left(a-3d\right)\left(a+3d\right)}{\left(a-d\right)\left(a+d\right)}=\frac{5}{6} \\ \Rightarrow \frac{a^2-9d^2}{a^2-d^2}=\frac{5}{6} \\ \Rightarrow \frac{{\left(14\right)}^2-9d^2}{{\left(14\right)}^2-d^2}=\frac{5}{6} \\ \Rightarrow \frac{196-9d^2}{196-d^2}=\frac{5}{6} \end{gathered}$$

$$\begin{gathered} \Rightarrow 1176-54d^2=980-5d^2 \\ \Rightarrow 196=49d^2 \\ \Rightarrow d^2=4 \\ \Rightarrow d=\pm 2 \end{gathered}$$

When d = 2, the terms of the AP are 8, 12, 16, 20. When d = −2, the terms of the AP are 20, 18, 12, 8.

Answer:

Let the first three numbers in an arithmetic progression be a − d, a, a + d.

The sum of the first three numbers in an arithmetic progression is 18.

a − d + a + a + d = 18
⇒ 3a = 18
⇒ a = 6

The product of the first and third term is 5 times the common difference.

$$\begin{gathered} \left(a-d\right)\left(a+d\right)=5d \\ \Rightarrow a^2-d^2=5d \\ \Rightarrow {\left(6\right)}^2-d^2=5d \\ \Rightarrow 36-d^2=5d \\ \Rightarrow d^2+5d-36=0 \\ \Rightarrow d^2+9d-4d-36=0 \\ \Rightarrow d\left(d+9\right)-4\left(d+9\right)=0 \\ \Rightarrow \left(d+9\right)\left(d-4\right)=0 \\ \Rightarrow d=4, -9 \\ \mathrm{For} d=4, \\ \mathrm{The} \mathrm{numbers} \mathrm{are} 6-4, 6, 6+4 \\ \mathrm{i}.\mathrm{e}. 2, 6, 10 \\ \mathrm{For} d=-9, \\ \mathrm{The} \mathrm{numbers} \mathrm{are} 6+9, 6, 6-9 \\ \mathrm{i}.\mathrm{e}. 15, 6, -3 \end{gathered}$$

Hence, the three numbers are 2, 6, 10 or 15, 6, −3.

Answer:

Here, we are given that the angles of a quadrilateral are in A.P, such that the common difference is 10°.

So, let us take the angles as

Now, we know that the sum of all angles of a quadrilateral is 360°. So, we get,

On further simplifying for a, we get,

So, the first angle is given by,

Second angle is given by,

Third angle is given by,

Fourth angle is given by,

Therefore, the four angles of the quadrilateral are .

Answer:

Suppose three parts of 207 are (a − d), a , (a + d) such that , (a + d)  >a >  (a − d).

$$\begin{gathered} a-d+a+a+d=207 \\ \Rightarrow 3a=207 \\ \Rightarrow a=69 \\ \mathrm{Now}, \left(a-d\right)\times a=4623 \\ \Rightarrow 69\left(69-d\right)=4623 \\ \Rightarrow \left(69-d\right)=67 \\ \Rightarrow d=2 \\ \mathrm{Therefore}, \mathrm{the} \mathrm{three} \mathrm{required} \mathrm{parts} \mathrm{are} 67, 69 \mathrm{and} 71. \end{gathered}$$

 

Answer:

Suppose the angles of a triangle are (a − d), a , (a + d) such that , (a + d)  >a >  (a − d).

$$\begin{gathered} a-d+a+a+d=180 \left[\mathrm{angle} \mathrm{sum} \mathrm{property}\right] \\ \Rightarrow 3a=180 \\ \Rightarrow a=60 \\ \mathrm{Now}, \left(a+d\right)=2\left(a-d\right) \\ \Rightarrow a+d=2a-2d \\ \Rightarrow a=3d \\ \Rightarrow d=\frac{60}{3}=20 \\ \mathrm{Therefore}, \mathrm{the} \mathrm{three} angles of a triangle \mathrm{are} 40, 60, 80. \end{gathered}$$

Answer:

Let the four terms of the AP be a − 3d, a − d, a + d and a + 3d.
Given:
(a − 3d) + (a − d) + (a + d) + (a + 3d) = 32
$\Rightarrow$ 4a = 32
$\Rightarrow$ a = 8

$$\begin{gathered} \mathrm{Also}, \\ \frac{\left(a-3d\right)\left(a+3d\right)}{\left(a-d\right)\left(a+d\right)}=\frac{7}{15} \\ \Rightarrow \frac{a^2-9d^2}{a^2-d^2}=\frac{7}{15} \\ \Rightarrow \frac{{\left(8\right)}^2-9d^2}{{\left(8\right)}^2-d^2}=\frac{7}{15} \\ \Rightarrow \frac{64-9d^2}{64-d^2}=\frac{7}{15} \end{gathered}$$

$$\begin{gathered} \Rightarrow 960-135d^2=448-7d^2 \\ \Rightarrow 512=128d^2 \\ \Rightarrow d^2=4 \\ \Rightarrow d=\pm 2 \end{gathered}$$
When a = 8 and d= 2, then the terms are 2, 6, 10, 14.
When a = 8 and d= −2, then the terms are 14, 10, 6, 2.

Answer:

In the given problem, we need to find the sum of terms for different arithmetic progressions. So, here we use the following formula for the sum of n terms of an A.P.,

Where; a = first term for the given A.P.

d = common difference of the given A.P.

n = number of terms

(i) To 10 terms

Common difference of the A.P. (d)

=

Number of terms (n) = 10

First term for the given A.P. (a) = 50

So, using the formula we get,

Therefore, the sum of first 10 terms for the given A.P. is.

(ii) To 12 terms.

Common difference of the A.P. (d)

=

Number of terms (n) = 12

First term for the given A.P. (a) = 1

So, using the formula we get,

Therefore, the sum of first 12 terms for the given A.P. is.

(iii) To 25 terms.

Common difference of the A.P. (d) =

Number of terms (n) = 25

First term for the given A.P. (a) = 3

So, using the formula we get,

On further simplifying, we get,

Therefore, the sum of first 25 terms for the given A.P. is.

(iv) To 12 terms.

Common difference of the A.P. (d) =

Number of terms (n) = 12

First term for the given A.P. (a) = 41

So, using the formula we get,

Therefore, the sum of first 12 terms for the given A.P. is.

(v) To 22 terms.

Common difference of the A.P. (d) =

Number of terms (n) = 22

First term for the given A.P. (a) =

So, using the formula we get,

Therefore, the sum of first 22 terms for the given A.P. is.

(vi) To n terms.

Common difference of the A.P. (d) =

Number of terms (n) = n

First term for the given A.P. (a) =

So, using the formula we get,

Now, taking 2 common from both the terms inside the bracket we get,

Therefore, the sum of first n terms for the given A.P. is

(vii) To n terms.

Number of terms (n) = n

First term for the given A.P. (a) =

Common difference of the A.P. (d) =

So, using the formula we get,

Now, on further solving the above equation we get,

Therefore, the sum of first n terms for the given A.P. is.

(viii) To 36 terms.

Common difference of the A.P. (d) =

Number of terms (n) = 36

First term for the given A.P. (a) = −26

So, using the formula we get,

Therefore, the sum of first 36 terms for the given A.P. is.

Answer:

Here, we are given the n^th term for various sequences. We need to find the first five terms of the sequence.

(i)

Here, the n^th term is given by the above expression. So, to find the first term we use, we get,

Similarly, we find the other four terms,

Second term (),

Third term,

Fourth term,

Fifth term,

Therefore, the first five terms for the given sequence are.

(ii)

Here, the n^th term is given by the above expression. So, to find the first term we use, , we get,

Similarly, we find the other four terms,

Second term (),

Third term,

Fourth term,

Fifth term,

Therefore, the first five terms for the given sequence are.

(iii)

Here, the n^th term is given by the above expression. So, to find the first term we use, we get,

Similarly, we find the other four terms,

Second term (),

Third term,

Fourth term,

Fifth term,

Therefore, the first five terms for the given sequence are.

(iv)

Here, the n^th term is given by the above expression. So, to find the first term we use, , we get,

Similarly, we find the other four terms,

Second term (),

Third term,

Fourth term,

Fifth term,

Therefore, the first five terms for the given sequence are.

(v)

Here, the n^th term is given by the above expression. So, to find the first term we use, we get,

Similarly, we find the other four terms,

Second term (),

Third term,

Fourth term,

Fifth term,

Therefore, the first five terms of the given A.P are.

(vi)

Here, the n^th term is given by the above expression. So, to find the first term we use , we get,

Similarly, we find the other four terms,

Second term (),

Third term,

Fourth term,

Fifth term,

Therefore, the first five terms for the given sequence are.

(vii)

Here, the n^th term is given by the above expression. So, to find the first term we use, we get,

Similarly, we find the other four terms,

Second term (),

Third term,

Fourth term,

Fifth term,

Therefore, the first five terms for the given sequence are.

(viii)

Here, the n^th term is given by the above expression. So, to find the first term we use, we get,

Similarly, we find the other four terms,

Second term (),

Third term,

Fourth term,

Fifth term,

Therefore, the first five terms for the given sequence are.

(ix)

Here, the n^th term is given by the above expression. So, to find the first term we use, we get,

Similarly, we find the other four terms,

Second term (),

Third term (),

Fourth term (),

Fifth term (),

Therefore, the first five terms of the given A.P are

Answer:

Here, we are given the n^th term for various sequences. We need to find the indicated terms of the A.P.

(i)

We need to find and

Now, to find term we use, we get,

Also, to find term we use, we get,

$$\begin{gathered} a_{15}=5\left(15\right)-4 \\ =75-4 \\ =71 \end{gathered}$$

Thus,

(ii)

We need to find and

Now, to find term we use, we get,

Also, to find term we use, we get,

Thus,

(iii)

We need to find and

Now, to find term we use, we get,

Also, to find term we use, we get,

Thus,

(iv)

We need to find, and

Now, to find term we use, we get,

Also, to find term we use, we get,

Similarly, to find term we use, we get,

Thus,

(v)

We need to find, and

Now, to find term we use, we get,