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Page No 74:

Question 1:

The common difference of the A.P. is 12q,1-2q2q,1-4q2q, ... is

(a) −1
(b) 1
(c) q
(d) 2q

Answer:

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

The given A.P. is 12q,1-2q2q,1-4q2q, ... 

Common difference = d = Second term − First term
                                       = 1-2q2q-12q
                                       = -2q2q=-1

Hence, the correct option is (a).

Page No 74:

Question 2:

The common difference of the A.P. 13,1-3b3,1-6b3, ... is

(a) 13
(b) -13
(c) −b
(d) b

Answer:

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

The given A.P. is 13,1-3b3,1-6b3, ...

Common difference = d = Second term − First term
                                       = 1-3b3-13
                                       = -3b3=-b

​Hence, the correct option is (c).

Page No 74:

Question 3:

The common difference of the A.P. 12b,1-6b2b,1-12b2b, ... is

(a) 2b
(b) −2b
(c) 3
(d) −3

Answer:

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

The given A.P. is 12b,1-6b2b,1-12b2b, ...

Common difference = d = Second term − First term
                                       = 1-6b2b-12b
                                       = -6b2b=-3

Hence, the correct option is (d).

Page No 74:

Question 4:

If k, 2k − 1 and 2k + 1 are three consecutive terms of an A.P., the value of k is

(a) −2
(b) 3
(c) −3
(d) 6

Answer:

Since, k, 2k − 1 and 2k + 1 are three consecutive terms of an A.P.

Then, Second term − First term = Third term − Second term = d (common difference)
⇒ 2k − 1 − k = 2k + 1 − (2k − 1)
⇒ k − 1 = 2k + 1 − 2k + 1
⇒ k − 1 = 2
⇒ k = 2 + 1
⇒ k = 3

Hence, the correct option is (b).

Page No 74:

Question 5:

The next term of the A.P. 7, 28, 63, ....

(a) 70
(b) 84
(c) 97
(d) 112

Answer:

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

The given A.P. is 7, 28, 63, ....
i.e.,  7, 4×7, 9×7, ....
i.e.,  7, 27, 37, ....

Common difference = d = Second term − First term
                                       = 27-7
                                       = 7

∴ Next term of the A.P. = 37+7
                                      = 47
                                      = 16×7
                                      = 112

​Hence, the correct option is (d).

Page No 74:

Question 6:

The first three terms of an A.P. respectively are 3y − 1, 3y + 5 and 5y + 1. Then, y equals

(a) −3
(b) 4
(c) 5
(d) 2
 

Answer:

​Since, 3y − 1, 3y + 5 and 5y + 1 are first three terms of an A.P.

Then, Second term − First term = Third term − Second term = d (common difference)
⇒ 3y + 5 − (3y − 1) = 5y + 1 − (3y + 5)
⇒ 3y + 5 − 3y + 1 = 5y + 1 − 3y − 5
⇒ 6 = 2y − 4
⇒ 2y = 6 + 4
⇒ 2y = 10
y = 5

Hence, the correct option is (c).

Page No 74:

Question 7:

If 1x+2,1x+3,1x+5 are in A.P. Then, x =

(a) 5

(b) 3

(c) 1

(d) 2

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,

Also,

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

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

So, the correct option is (c).



Page No 75:

Question 8:

The nth term of an A.P., the sum of whose n terms is Sn, is

(a) Sn + Sn−1

(b) Sn − Sn−1

(c) Sn + Sn+1

(d) Sn − Sn+1

Answer:

A.P. we use following formula,

So, the nth term of the A.P. is given by . Therefore, the correct option is (b).

Page No 75:

Question 9:

The common difference of an A.P., the sum of whose n terms is Sn, is

(a) Sn − 2Sn−1 + Sn−2

(b) Sn − 2Sn−1 − Sn−2

(c) Sn − Sn−2

(d) Sn − Sn−1

Answer:

Here, we are given an A.P. the sum of whose n terms is Sn. So, to calculate the common difference of the A.P, we find two consecutive terms of the A.P.

Now, the nth term of the A.P will be given by the following formula,

Next, we find the (n − 1)th term using the same formula,

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

Therefore,

Hence the correct option is (a).

Page No 75:

Question 10:

The sum of first 20 odd natural numbers is

(a) 100
(b) 210
(c) 400
(d) 420

Answer:

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

We know that, sum of first n terms = Sn n2[2a + (n − 1)d]

The given series is 1 + 3 + 5 + ......

First term = =  1.

Common difference = d = 3 − 1 = 2

∴ S20 202[2 × 1 + (20 − 1)2]
          = 10(2 + 19 × 2)
          = 10(40)
         = 400

Hence, the correct option is (c).

Page No 75:

Question 11:

If 18, a, b, −3 are in A.P., the a + b =

(a) 19

(b) 7

(c) 11

(d) 15

Answer:

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

First term (a1) =

Second term (a2) =

Third term (a3) =

Fourth term (a4)=

So, in an A.P. the difference of two adjacent terms is always constant. So, we get,

Also,

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

Therefore,

Hence the correct option is (d).

Page No 75:

Question 12:

The first term of an A.P. is p and the common difference is q, then its 10th term is
(a) q + 9p
(b) p – 9q
(c) p + 9q
(d) 2p + 9q

Answer:

The nth term of an AP = a+n-1d, where a and d are the first term and common difference respectively.
Therefore, 10th term = p+10-1q = p+9q.

Hence, the correct answer is option (c).

Page No 75:

Question 13:

The value of x for which 2x, x + 10 and 3x + 2 are the three consecutive terms of an A.P; is
(a) 6
(b) –6
(c) 18
(d) –18

Answer:

Given 2x, x+10, 3x+2 are the consecutive terms of an AP.
Therefore, the common difference will be same.
x+10-2x=3x+2-x+10x+10-2x=3x+2-x-1010-x=2x-83x=18x=6
Hence, the correct answer is option (a).

Page No 75:

Question 14:

If the sum of three consecutive terms of an increasing A.P. is 51 and the product of the first and third of these terms is 273, then the third term is

(a) 13

(b) 9

(c) 21

(d) 17

Answer:

In the given problem, the sum of three consecutive terms of an A.P is 51 and the product of the first and the third terms is 273.

We need to find the third term.

Here,

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

So,

Also,

Further solving for d,

 

Now, it is given that this is an increasing A.P. so d cannot be negative.

So, d = 4

Substituting the values of a and d in the expression for the third term, we get,

Third term =

So,

Therefore, the third term is

Hence, the correct option is (c).

Page No 75:

Question 15:

3-1n+3-2n+3-3n+... upto n terms is

(a) 123n-1

(b) 123n+1

(c) 125n-1

(d) 125n+1

Answer:

Given that, 3-1n+3-2n+3-3n+... upto n terms.
So, the sum upto n terms is given as: 3-1n+3-2n+3-3n+...+3-nn.

Thus, it is an AP with
First term, a=3-1n
Common difference, d=3-2n-3-1n=-1n

Therefore,
Sn=3-1n+3-2n+3-3n+...+3-nn=3+3+...+3n times-1n1+2+...+n=3×n-1nnn+12                         sum of first n natural numbers=nn+12=3n-n+12=6n-n-12=5n-12

Hence, the correct answer is option (c).

Page No 75:

Question 16:

The sum of first 16 terms of the A.P.: 10, 6, 2, ..., is
(a) –320
(b) 320
(c) –352
(d) –400

Answer:

The given A.P. is:
10, 6, 2, ...,

First term = a = 10
Common difference = d = 6 − 10 = −4

S16=1622a+16-1d    =82a+15d    =8210+15-4    =820-60    =8×-40    =-320

​Hence, the correct option is (a).

Page No 75:

Question 17:

If the first term of an A.P. is –5 and the common difference is 2, then the sum of first 6 terms is
(a) 0
(b) 5
(c) 6
(d) 15

Answer:

Given:
First term = a = −5
Common difference = d = 2

S6=622a+6-1d    =32a+5d    =32-5+52    =3-10+10    =3×0    =0

​Hence, the correct option is (a).

Page No 75:

Question 18:

The 4th term from the end of the AP: –11, –8, –5, ..., 49 is
(a) 37
(b) 40
(c) 43
(d) 58

Answer:

The given A.P. is: –11, –8, –5, ..., 49

Rearranging the terms from last to first, we get
49, 46, 43, ..... –11

First term = a = 49
Common difference = d = 46 – 49
                                       =  –3

The fourth term is:
a4=a+4-1d    =49+3-3    =49-9    =40

Therefore, the 4th term from the end of the AP: –11, –8, –5, ..., 49 is 40.

​Hence, the correct option is (b).

Page No 75:

Question 19:

Which term of the A.P. 21, 42, 63, 84, ... is 210?
(a) 9th
(b) 10th
(c) 11th
(d) 12th

Answer:

The given A.P. is: 21, 42, 63, 84, ...

First term = a = 21
Common difference = d = 42 – 21
                                       =  21
nth term = an = 210

Now,
an=a+n-1d210=21+n-121210=21+21n-21210=21nn=10

Therefore, the 10th term of the A.P. 21, 42, 63, 84, ... is 210.

​Hence, the correct option is (b).

Page No 75:

Question 20:

If the 2nd term of an A.P, is 13 and 5th term is 25, what is its 7th term?
(a) 30
(b) 33
(c) 37
(d) 38

Answer:

Given:
2nd term = a2 = 13
5th term = a5 = 25

Now,
an=a+n-1da2=a+2-1d13=a+da+d=13     ...1a5=a+5-1d25=a+4da+4d=25     ...2Subtracting 1 from 2, we get3d=12d=4Substituting the value of d in 1, we geta=9Therefore, a=9 and d=4.Thus,a7=a+7-1d    =9+64    =33

Therefore, the 7th term is 33.

​Hence, the correct option is (b).

Page No 75:

Question 21:

The first and last terms of an A.P. are 1 and 11. If the sum of its terms is 36, then the number of terms will be

(a) 5

(b) 6

(c) 7

(d) 8

Answer:

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

First term (a) = 1

Last term (an) = 11

Sum of its terms

Now, as we know,

Where, a = the first term

l = the last term

So, we get,

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

Hence the correct option is (b).

Page No 75:

Question 22:

If four numbers in A.P. are such that their sum is 50 and the greatest number is 4 times, the least, then the numbers are

(a) 5, 10, 15, 20

(b) 4, 101, 16, 22

(c) 3, 7, 11, 15

(d) none of these

Answer:

Here, we are given that four numbers are in A.P., such that their 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,

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

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

Hence, the correct option is (a).

Page No 75:

Question 23:

If the first term of an A.P. is 2 and common difference is 4, then the sum of its 40 terms is 
(a) 3200
(b) 1600
(c) 200
(d) 2800

Answer:

In the given problem, we need to find the sum of 40 terms of an arithmetic progression, where we are given the first term and the common difference. 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

Given,

First term (a) = 2

Common difference (d) = 4

Number of terms (n) = 40

So, using the formula we get,

Therefore, the sum of first 40 terms for the given A.P. is. So, the correct option is (a).

Page No 75:

Question 24:

The number of terms of the A.P. 3, 7, 11, 15, ... to be taken so that the sum is 406 is

(a) 5

(b) 10

(c) 12

(d) 14

(e) 20

Answer:

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

Here, we need to find the number of terms n such that the sum of n terms is 406.

So here, we will use the formula,

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

d = common difference of the given A.P.

n = number of terms

The first term (a) = 3

The sum of n terms (Sn) = 406

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

So, on substituting the values in the formula for the sum of n terms of an A.P., we get,

So, we get the following quadratic equation,

On solving by splitting the middle term, we get,

Further,

Or,

Since, the number of terms cannot be a fraction, the number of terms (n) is

Hence, the correct option is (d).

Page No 75:

Question 25:

Sum of n terms of the series 2+8+18+32+..is

(a) n (n + 1) 2

(b) 2n (n + 1) 

(c) n (n+1)2

(d) 1

Answer:

In the given problem, we need to find the sum of terms for a given arithmetic progression,

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

Here,

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 common from both the terms inside the bracket we get,

Therefore, the sum of first n terms for the given A.P. is. So, the correct option is (c).



Page No 76:

Question 26:

The 9th term of an A.P. is 449 and 449th term is 9. The term which is equal to zero is

(a) 501th

(b) 502th

(c) 508th

(d) none of these

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 n

Also, we know,

For the 9th term (n = 9),

Similarly, for the 449th term (n = 449),

Subtracting (3) from (4), we get,

Now, to find a, we substitute the value of d in (3),

So, for the given A.P

So, let us take the term equal to zero as the nth term. So,

So,

Therefore, the correct option is (d).

Page No 76:

Question 27:

If the first term of an A.P. is a and nth term is b, then its common difference is

(a) b-an+1

(b) b-an-1
 
(c) b-an

(d) b+an-1

Answer:

Here, we are given the first term of the A.P. as a and the nth term (an) as b. So, let us take the common difference of the A.P. as d.

Now, as we know,

On substituting the values given in the question, we get.

Therefore,

Hence the correct option is (b).

Page No 76:

Question 28:

If 5+9+13+... to n terms7+9+11+...to (n+1) terms=1716, then n =

(a) 8

(b) 7

(c) 10

(d) 11

Answer:

Here, we are given,

We need to find n.

So, first let us find out the sum of n terms of the A.P. given in the numerator. 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

Here,

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,

Similarly, we find out the sum of terms of the A.P. given in the denominator.

Here,

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 substituting the values of (2) and (3) in equation (1), we get,

Further solving the quadratic equation for n by splitting the middle term, we get,

So, we get

Or

Since n is a whole number, it cannot be a fraction. So,

Therefore, the correct option is (b).

Page No 76:

Question 29:

The sum of n terms of an A.P. is 3n2 + 5n, then 164 is its

(a) 24th term

(b) 27th term

(c) 26th term

(d) 25th term

Answer:

Here, the sum of first n terms is given by the expression,

We need to find which term of the A.P. is 164.

Let us take 164 as the nth term.

So we know that the nthterm of an A.P. is given by,

So,

Using the property,

We get,

Further solving for n, we get

Therefore,

Hence the correct option is (b).

Page No 76:

Question 30:

If the nth term of an A.P. is 2n + 1, then the sum of first n terms of the A.P. is

(a) n(n − 2)

(b) n(n + 2)

(c) n(n + 1)
 
(d) n(n − 1)

Answer:

Here, we are given an A.P. whose nth term is given by the following expression, . We need to find the sum of first n terms.

So, here we can find the sum of the n terms of the given A.P., using the formula,

Where, a = the first term

l = the last term

So, for the given A.P,

The first term (a) will be calculated using in the given equation for nth term of A.P.

Now, the last term (l) or the nth term is given

So, on substituting the values in the formula for the sum of n terms of an A.P., we get,

Therefore, the sum of the n terms of the given A.P. is. So the correct option is (b).

Page No 76:

Question 31:

The sum of first 24 terms of the sequence whose nth terms is given by an=3+23n
(a) 270
(b) 272
(c) 382
(d) 384

Answer:

Given that, an=3+23n.

Thus,
an+1=3+23n+1=3+23n+23=113+23n

So,
d=an+1-an=113+23n-3-23n=23

Now,
a=a1=3+231=113

The sum of n terms of an AP is given as: Sn=n22a+n-1d.

S24=2422×113+24-123=1222+463=4×68=272

Hence, the correct answer is option (b).

Page No 76:

Question 32:

The sum of first five multiples of 3 is
(a) 45
(b) 55
(c) 65
(d) 75

Answer:

The multiples of 3 form an AP with common difference, d = 3 and first term, a = 3.

Here, n = 5.

Therefore,
S5=522×3+5-13                         Sn=n22a+n-1d=526+12=5×182=45

Hence, the correct answer is option (a).

Page No 76:

Question 33:

If the sum of P terms of an A.P. is q and the sum of q terms is p, then the sum of p + q terms will be

(a) 0

(b) pq

(c) p + q

(d) −(p + q)

Answer:

In the given problem, we are given and

We need to find

Now, as we know,

So,

Similarly,

Subtracting (2) from (1), we get

Now,

Thus,

Hence, the correct option is (d).

Page No 76:

Question 34:

If the sum of n terms of an A.P. be 3n2 + n and its common difference is 6, then its first term is

(a) 2

(b) 3

(c) 1

(d) 4

Answer:

In the given problem, the sum of n terms of an A.P. is given by the expression,

Here, we can find the first term by substitutingas sum of first term of the A.P. will be the same as the first term. So we get,

Therefore, the first term of this A.P is. So, the correct option is (d).

Page No 76:

Question 35:

Two A.P.'s have the same common difference. The first term of one of these is 8 and that of the other is 3. The difference between their 30th term is

(a) 11

(b) 3

(c) 8

(d) 5

Answer:

Here, we are given two A.P.’s with same common difference. Let us take the common difference as d.

Given,

First term of first A.P. (a) = 8

First term of second A.P. (a) = 3

We need to find the difference between their 30th terms.

So, let us first find the 30th term of first A.P.

Similarly, we find the 30th term of second A.P.

Now, the difference between the 30th terms is,

Therefore,

Hence, the correct option is (d).

Page No 76:

Question 36:

Let Sn denote the sum of n terms of an A.P. whose first term is a. If the common difference d is given by = Sn − kSn−1 + Sn−2, then k =

(a) 1

(b) 2

(c) 3

(d) none of these.
 

Answer:

In the given problem, we are given

We need to find the value of k

So here,

First term = a

Common difference = d

Sum of n terms = Sn

Now, as we know,

Also, for n-1 terms,

Further, for n-2 terms,

Now, we are given,

Using (1), (2) and (3) in the given equation, we get

Taking common, we get,

Taking 2 common from the numerator, we get,

Therefore,

Hence, the correct option is (b).

Page No 76:

Question 37:

The first and last term of an A.P. are a and l respectively. If S is the sum of all the terms of the A.P. and the common difference is given by l2-a2k-(l+a), then k =

Answer:

In the given problem, we are given the first, last term, sum and the common difference of an A.P.

We need to find the value of k

Here,

First term = a

Last term = l

Sum of all the terms = S

Common difference (d) =

Now, as we know,

Further, substituting (1) in the given equation, we get

Now, taking d in common, we get,

Taking (n-1) as common, we get,

Further, multiplying and dividing the right hand side by 2, we get,

Now, as we know,

Thus,

Therefore, the correct option is (b).

Page No 76:

Question 38:

If the sum of first n even natural numbers is equal to k times the sum of first n odd natural numbers, then k =

(a) 1n

(b) n-1n

(c) n+12n

(d) n+1n

Answer:

In the given problem, we are given that the sum of the first n even natural numbers is equal to k times the sum of first n odd natural numbers.

We need to find the value of k

Now, we know that the first odd natural number is 1. Also, all the odd terms will form an A.P. with the common difference of 2.

So here,

First term (a) = 1

Common difference (d) = 2

So, let us take the number of terms as n

Now, as we know,

So, for n terms,

Also, we know that the first even natural number is 2. Also, all the odd terms will form an A.P. with the common difference of 2.

So here,

First term (a) = 2

Common difference (d) = 2

So, let us take the number of terms as n

So, for n terms,

Solving further, we get

Now, as the sum of the first n even natural numbers is equal to k times the sum of first n odd natural numbers

Using (1) and (2), we get

Therefore,

Hence, the correct option is (d).

Page No 76:

Question 39:

If the first, second and last term of an A.P. are a, b and 2a respectively, its sum is

(a) ab2(b-a)

(b) ab(b-a)

(c) 3ab2(b-a)

(d) none of these

Answer:

In the given problem, we are given first, second and last term of an A.P. We need to find its sum.

So, here

First term = a

Second term (a2) = b

Last term (l) = 2a

Now, using the formula

Also,

Further as we know,

Substituting (2) in the above equation, we get

 

Using (1), we get

Thus,

Therefore, the correct option is (c).

Page No 76:

Question 40:

If S1 is the sum of an arithmetic progression of 'n' odd number of terms and S2 the sum of the terms of the series in odd places, then S1S2=

(a) 2nn+1

(b) nn+1

(c) n+12n

(d) n+1n

Answer:

In the given problem, we are given as the sum of an A.P of ‘n’ odd number of terms and the sum of the terms of the series in odd places.

We need to find

Now, let a1, a2…. an be the n terms of A.P

Where n is odd

Let d be the common difference of the A.P

Then,

And be the sum of the terms of the places in odd places,

Where, number of terms =

Common difference = 2d

So,

Now,

Thus,

Therefore, the correct option is (a).

Page No 76:

Question 41:

If in an A.P. Sn = n2p and Sm = m2p, where Sr denotes the sum of r terms of the A.P., then Sp is equal to

(a) 12p3

(b) m n p

(c) p3

(d) (m + n) p2

Answer:

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

We need to find

Now, as we know,

Where, first term = a

Common difference = d

Number of terms = n

So,

Similarly,

Equating (1) and (2), we get,
12n2a+nd-d=12m2a+md-d
m2a+nd-d=n2a+md-d
2am+mnd-md=2an+mnd-nd

Solving further, we get,

Further, substituting (3) in (1), we get,

Now,

Thus,

Therefore, the correct option is (C).



Page No 77:

Question 42:

If Sn denote the sum of the first n terms of an A.P. If S2n = 3Sn, then S3n : Sn is equal to

(a) 4

(b) 6

(c) 8

(d) 10

Answer:

Here, we are given an A.P. whose sum of n terms is Sn and.

We need to find.

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

So, first we find S3n,

Similarly,

Also,

Now,

So, using (2) and (3), we get,

On further solving, we get,

So,

Taking common, we get,

Therefore,

Hence, the correct option is (b).

Page No 77:

Question 43:

In an AP. Sp = q, Sq = p and Sr denotes the sum of first r terms. Then, Sp+q is equal to

(a) 0

(b) −(p + q)

(c) p + q

(d) pq

Answer:

In the given problem, we are given and

We need to find

Now, as we know,

So,

Similarly,

Subtracting (2) from (1), we get

Now,

Thus,

Hence, the correct option is (b).

Page No 77:

Question 44:

If Sr denotes the sum of the first r terms of an A.P. Then , S3n: (S2nSn) is

(a) n

(b) 3n

(c) 3

(d) none of these

Answer:

Here, we are given an A.P. whose sum of r terms is Sr. We need to find.

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

So, first we find S3n,

Similarly,

Also,

So, using (1), (2) and (3), we get,

Taking common, we get,

Therefore,

Hence, the correct option is (c).

Page No 77:

Question 45:

If the sums of n terms of two arithmetic progressions are in the ratio 3n+55n-7, then their nth terms are in the ratio

(a) 3n-15n-1

(b) 3n+15n+1

(c) 5n+13n+1

(d) 5n-13n-1

Answer:

In the given problem, the ratio of the sum of n terms of two A.P’s is given by the expression,

We need to find the ratio of their nth terms.

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

So,

Where, a and d are the first term and the common difference of the first A.P.

Similarly,

Where, a and d are the first term and the common difference of the first A.P.

So,

Equating (1) and (2), we get,

Now, to find the ratio of the nth term, we replace n by. We get,

As we know,

Therefore, we get,

Hence the correct option is (b).

Page No 77:

Question 46:

If Sn denote the sum of n terms of an A.P. with first term a and common difference d such that SxSkx is independent of x, then

(a) d= a

(b) d = 2a

(c) a = 2d

(d) d = −a

Answer:

Here, we are given an A.P. with a as the first term and d as the common difference. The sum of n terms of the A.P. is given by Sn.

We need to find the relation between a and d such thatis independent of

So, let us first find the values of Sx and Skx using 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

So, we get,

Similarly,

So,

Now, to get a term independent of x we have to eliminate the other terms, so we get

So, if we substitute, we get,

Therefore,

Hence, the correct option is (b).

Page No 77:

Question 47:

The sum of n terms of two A.P.'s are in the ratio 5n + 9 : 9n + 6. Then, the ratio of their 18th term is

(a) 179321

(b) 178321

(c) 175321

(d) 176321

Answer:

In the given problem, the ratio of the sum of n terms of two A.P’s is given by the expression,

We need to find the ratio of their 18th terms.

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

So,

Where, a and d are the first term and the common difference of the first A.P.

Similarly,

Where, a’ and d are the first term and the common difference of the first A.P.

So,

Equating (1) and (2), we get,

Now, to find the ratio of the nth term, we replace n by. We get,

As we know,

Therefore, for the 18th terms, we get,

Hence

Hence no option is correct.

Page No 77:

Question 48:

Lumber is a significant natural resource that contributes jobs to the US economy. Lumber companies source their raw materials from privately-managed or government-leased forests. In order to process tree wood into usable lumber, this raw material is transported to lumber mills, where it is cut to different sizes. Lumber is primarily used by the construction industry, though it can also be used to produce furniture, paper and pulp, and composites such as plywood. A lumber company stacks 200 logs in the following manner:



20 logs in the bottom row, 19 in the next row, 18 in the row next to it and so on as shown in the given figure. Based on the above information answer the following questions.
(i) Number of logs in first row, second row, third row, …..

(a) follow a pattern forming an A.P. with common difference 1.
(b) follow a pattern forming an A.P. with common difference –1.
(c) do not follow any specific pattern.
(d) follow a pattern forming an A.P. with common difference 2.
 
(ii) The number of rows in which 200 logs are stacked is
(a) 25
(b) 20
(c) 16
(d) 10
 
(iii) The number of logs in the top row is
(a) 5
(b) 7
(c) 10
(d) 2
 
(iv) The number of logs in the middle rows are:
(a) 11, 10
(b) 12, 11
(c) 14, 13
(d) 13, 12
 
(v) The number of logs in the top two rows is
(a) 10
(b) 11
(c) 9
(d) 12

Answer:

(i) Given that, there are 20 logs in the bottom row, 19 in the next row, 18 in the row next to it and so on.
Thus, the number of logs in first row, second row, third row, … follow a pattern forming an A.P. with common difference –1 as the number of logs decrease by 1 with every other row.

Hence, the correct answer is option (b).

(ii) Given that, the total number of logs is 200. Let the total number of rows be n.
200=n22×20+n-1-1                          Sn=n22a+n-1d200=n240-n+1400=41n-n2n2-41n+400=0n2-25n-16n+400=0n-25n-16=0n=25 or n=16

Now,
a16=20+16-1-1                          an=a+n-1da16=20-15a16=5

And
a25=20+25-1-1                          an=a+n-1da25=20-24a25=-4, which is not possible as it represents the number of logs in that row

Therefore, the number of rows in which the logs are placed is 16.

Hence, the correct answer is option (c).

(iii) There are a total of 16 rows in which 200 logs are arranged. And the number of logs in the top row or in the 16th row is 5.
As,
a16=20+16-1-1                          an=a+n-1da16=20-15a16=5

Hence, the correct answer is option (a).

(iv) Total number of rows = 16.
So, the middle rows in the given arrangement are rows 8th and 9th.
a8=20+8-1-1                          an=a+n-1da8=20-7a8=13

And
a9=20+9-1-1                          an=a+n-1da9=20-8a9=12

Hence, the correct answer is option (d).

(v)  The number of logs in the top row i.e., in the 16th row is 5.
And in the second top row i.e., in the 15th row = 6.  (∵ d = − 1 and a15 = a16d)
Thus, the top two rows have 6 logs and 5 logs. 
So, total number of logs = 6 + 5
                                        = 11

Hence, the correct answer is option (b).



Page No 78:

Question 49:

A small terrace at a football ground comprises of 15 steps each of which is 50 m long and built of solid concrete. Each step rises of 14 m and a tread of 12 m (see in the given figure). Let V1, V2, V3, ... , V15 respectively denote the volumes of concrete required to build the first, second, third, ..., fifteenth step. Based on the above information, answer following questions:



(i) Heights of first, second, third, ..., 15th steps form

(a) an A.P. with common difference 14 m

(b) an A.P. with common difference 12 m

(c) an A.P. with common difference 34 m

(d) an A.P. with common difference -14 m
 
(ii) The value of V2 is
(a) 25 m3
(b) 50 m3
(c) 12.5 m3
(d) 6.25 m3

(iii) The volume of concrete used in the middle step is
(a) 25 m3
(b) 50 m3
(c) 75 m3
(d) 6.25 m3

(iv) The sum of the surface areas of 15 treads is
(a) 350 m2
(b) 400 m2
(c) 375 m2
(d) 475 m2

(v) The total volume of the concrete required to build the terrace is
(a) 800 m3
(b) 375 m3
(c) 650 m3
(d) 750 m3

Answer:

(i) Given that, the ground comprises of 15 steps each of which is 50 m long and each step rises of 14 m and a tread has a constant length of 12 m. Since the height of each step increases by 14 m, it forms an A.P. with common difference 14 m.

Hence, the correct answer is option (a).

(ii) The height of each step rises by 14 m, thus forming the AP as 14,12,34,1,... with d14 and a = 14.
The volume of the second step is equal to the volume of cuboid with dimensions 12 m, 12 m and 50 m.
Thus,
 Volume of second step,V2=12×12×50                Volume of cuboid=lbh=252 m3=12.5 m3

Hence, the correct answer is option (c).

(iii) Since there are 15 steps, the middle one is the 8th one. The height of the middle step is obtained as follows:
a8=a+8-1d                              an=a+n-1d=14+7×14=8×14=2

The volume of the middle step is equal to the volume of cuboid with dimensions 2 m, 12 m and 50 m.
Thus,
Volume of concrete used in middle step=2×12×50           Volume of cuboid=lbh=50 m3

Hence, the correct answer is option (b).

(iv) The sum of the surface areas of 15 treads is equal to a rectangle with breadth 50 m and length formed by 15 steps of 12 m each.
Thus,
Surface Area of 15 treads=50×12×15           SA of rectangle=lb=375 m2

Hence, the correct answer is option (c).

(v) Volume of concrete in 1st step = 50 m ×12 m × 14 m = 6.25 m³
Volume of concrete used in 2nd step = 50 m × 12 m × 12 m = 12.50 m³
Volume of concrete used in 3rd step = 50 m × 12 m × 34 m = 18.75 m³

It can be observed that the volumes of concrete in these steps are in an A.P.: 6.25 m³, 12.50 m³, 18.75 m³, .... with first term a = 6.25 and common difference d = 6.25.

Let nth term of AP be the 15 as the number of steps is 15.

S15=1522×6.25+15-1×6.25                                   Sn=n22a+n-1d=15212.5+87.5=152×100=750 m3

Hence, the correct answer is option (d).

Page No 78:

Question 50:

A carpenter wants to manufacture a 3 metre ladder having rungs 25 cm apart (see in the given figure). The rungs decrease uniformly in length from 45 cm at the bottom to 25 cm at the top and the top and bottom rungs are 2.5 metre apart.

Based on the above information, answer the following questions.
(i) Total number of rungs in the ladder is

(a) 10
(b) 9
(c) 11
(d) 12

(ii) The lengths of rungs from bottom to top form an A.P. with first and last terms as 45 cm and 25 cm respectively. The common difference of the A.P. formed is
(a) –2 cm
(b) –2.5 cm
(c) 4.5 cm
(d) 2 cm

(iii) The length of the middle rung is
(a) 33 cm
(b) 35 cm
(c) 37 cm
(d) 35.5 cm

(iv) Length of the wood used for rungs is
(a) 3.75 metres
(b) 2.85 metres
(c) 3.85 metres
(d) 4 metres

(v) Length of the wood required for the ladder
(a) 68.5 metres
(b) 9.85 metres
(c) 5.85 metres
(d) 8.85 metres

(vi) If the wood costs ₹100 per metres, the cost of ladder is
(a) ₹685
(b) ₹585
(c) ₹885
(d) ₹985

Answer:

(i) Given that, the first and the last rungs are 2.5 m apart and the distance between two rungs is 25 cm. The total number of rungs in the ladder is 1 more than the number of gaps in the ladder of 2.5 m with each rung 25 cm apart.
Thus,
Total number of rungs in the ladder=2.5 m25 cm+1=25025+1=10+1=11

Hence, the correct answer is option (c).

(ii) The lengths of rungs from bottom to top form an A.P. with first and last terms as 45 cm and 25 cm respectively.
Thus,
a11=a1+11-1d                         an=a1+n-1d25=45+10d10d=-20d=-2

Hence, the correct answer is option (a).

(iii) Out of 11 rungs, the middle one is the 6th rung. We have to find a6.
Now,
a6=a1+6-1d                         an=a1+n-1da6=45+5-2a6=45-10a6=35

Therefore, the length of the middle rung is 35 cm.

Hence, the correct answer is option (b).

(iv) It is given that the rungs are 25 cm apart and the top and bottom rungs are 2.5 m apart. Now, as the lengths of the rungs decrease uniformly, they will be in an A.P. The length of the wood required for the rungs equals the sum of all the terms of this A.P. with first term, = 45, last term, l = 25 and n = 11.

Therefore,
S11=1122×45+11-1-2                     Sn=n22a+n-1d=11290-20=11×35=385

Therefore, the length of the wood required for the rungs is 385 cm or 3.85 m.

Hence, the correct answer is option (c).

(v) The length of the wood required for the rungs is 3.85 m. For the vertical parts of the ladder, two woods of 3 m each are required.
Thus,
Total length of the wood required for the ladder = 3.85 + 3 + 3
                                                                             = 9.85 m

Hence, the correct answer is option (b).

(vi) Given that, the wood costs ₹100 per metres.
Thus,
Cost of wood required for the ladder = 100×9.85
                                                            = ₹985

Hence, the correct answer is option (d).



Page No 79:

Question 51:

In a potato race, a bucket is placed at the starting point, which is 5 m from the first potato, and the other potatoes are placed 3 m apart in a straight line. There are n potatoes in the line (See in the given figure). Each competitor starts from the bucket, picks up the nearest potato, runs back with it, drops it in the bucket, runs back to pick up the next potato, runs to the bucket to drop it in the bucket, and she continues in the same way until all the potatoes are in the bucket. Based on the above information, answer the following questions:


 
(i) Distance run by the competitor to pickup and drop first potato in the bucket is

(a) 5 cm
(b) 8 m
(c) 10 m
(d) 7 m
 
(ii) Distance run by the competitor to pick up and drop nth potato in the bucket is
(a) (3n + 2) m
(b) 2(3n + 2) m
(c) 2(3n – 1) m
(d) 6(n – 1) m
 
(iii) Total distance run by the competitor to pick up and drop first four potatoes is
(a) 36 metres
(b) 40 metres
(c) 86 metres
(d) 76 metres
 
(iv) Total distance run by the competitor to pick up and drop n potatoes in the basket is
(a) n(3n + 2) m
(b) 2n(3n + 2) m
(c) n(3n + 7) m
(d) 3n2 + 7 m
 
(v) If d1, d2, d3,..., dn denote distances run by the competitor to pick up first, second, third, …, nth potato respectively, then d1, d2, d3,..., dn form an A.P. with common difference
(a) 3
(b) 6
(c) 7
(d) 5

Answer:

(i) Distance run by the competitor to pickup and drop first potato in the bucket is = 5 × 2
                                                                                                                                  = 10 m

Hence, the correct answer is option (c).

(ii) The distances of the potatoes from the bucket form an AP as follows: 5, 8, 11, 14, 17, ... with common difference d = 3 and a1=5.
The distance run by the competitor to pick up and drop nth potato in the bucket is double the distance of the nth potato from the bucket.
The distance of the nth potato from the bucket is given by an.
Here,
an=a1+n-1d=5+n-13=2+3n

Total distance run by the competitor to pick up and drop nth potato in the bucket = 2(2 + 3n) m

Hence, the correct answer is option (b).

(iii) The distances covered to pick up and drop potatoes form an AP as follows: 5 × 2, 8 × 2, 11 × 2, 14 × 2, 17 × 2, ...
i.e., 10, 16, 22, 28, 34, ... with a1=10 and d = 6.
Thus,
S4=422×10+4-16                        Sn=n22a1+n-1d=220+18=76 m

Hence, the correct answer is option (d).

(iv) The distance covered to pick up and drop potatoes forms an AP as follows: 10, 16, 22, 28, 34, ... with a1=10 and d = 6.
Thus,
Sn=n22×10+n-16                        Sn=n22a1+n-1d=n214+6n=n7+3n m

Hence, the correct answer is option (c).

(v) Let d1d2d3,..., dn denote distances run by the competitor to pick up first, second, third, …, nth potato respectively. The distance covered to pick up and drop potatoes forms an AP as follows: 10, 16, 22, 28, 34, ... with a1=10 and d = 6. Then, d1d2d3,..., dn form an A.P. with common difference 6.

Hence, the correct answer is option (b).

 



Page No 80:

Question 52:

In a lemon race, a bucket is placed at a starting point, which is 6 m away from the first lemon and other lemons are placed 4 m apart from each other in a straight line. There are 10 lemons in a line. Riya starts from the bucket, picks up the nearest lemon, runs back with it, drops it in the bucket, runs back to pick up the next lemon, runs to the bucket to drop it in and continues until all the lemons are in the bucket. Based on the above information, answer the following questions:



(i) The lemons are placed in a straight line depicts which part of sequence?

(a) Geometric
(b) Arithmetic
(c) Linear
(d) Harmonic

(ii) The total distance covered by Riya is
(a) 370 m
(b) 480 m
(c) 460 m
(d) 400 m

(iii) The formula to find nth term of the Arithmetic sequence (progression) is
(a) an = a – (n – 1)d
(b) an = a (n – 1)d
(c) an = a + (n – 1)d
(d) Sn=n22a+n1d

(iv) The difference between the terms of arithmetic sequence is called as
(a) common ratio
(b) common difference
(c) common term
(d) none of these

Answer:

(i) The distances of the lemons from the bucket form an AP as follows: 6, 10, 14, 18, 22, ... with common difference d = 4 and a1 = 6.

Hence, the correct answer is option (b).

(ii) The distances covered to pick up and drop lemons form an AP as follows: 6 × 2, 10 × 2, 14 × 2, 18 × 2, 22 × 2, ...
i.e., 12, 20, 28, 36, 44, ... with a1=12 and d = 8.
Thus,
S10=1022×12+10-18                        Sn=n22a1+n-1d=524+72=5×96=480 m

Hence, the correct answer is option (b).

(iii) The formula to find nth term of the Arithmetic sequence (progression) is an = a + (n – 1)d.

Hence, the correct answer is option (c).

(iv) ​The difference between the terms of arithmetic sequence is known as the common difference.

Hence, the correct answer is option (b).

Page No 80:

Question 53:

The given figure shows playing cards stacked together in a following manner:
56 cards are stacked in this manner. 14 cards are in the bottom row, 12 in the next row, 10 in the row next to it and so on. Based on this information, answer the following questions:


 
(i) The total number of rows in which the cards are stacked is

(a) 7
(b) 6
(c) 8
(d) 9

(ii) The number of cards in the top row is
(a) 4
(b) 6
(c) 1
(d) 2

(iii) The mathematical concept applied in solving the above problem is
(a) Linear equations
(b) Probability
(c) Arithmetic progression
(d) Coordinate geometry

Answer:


(i) There are a total of 56 cards placed in such a fashion: 14 cards are in the bottom row, 12 in the next row, 10 in the row next to it and so on. Thus, it forms an AP with common difference d = −2 and a=a1=14.
56=n22×14+n-1-2                            Sn=n22a+n-1d112=28n-2n2+2n2n2-30n+112=0n2-15n+56=0n-7n-8=0n=7 or n=8

Now,
a7=14+7-1-2                            an=a+n-1d=14-12=2
And,
a8=14+8-1-2                            an=a+n-1d=14-14=0

Since the number of cards in the 8th row is 0, the total number of rows in which the cards are placed is 7.

Hence, the correct answer is option (a).

(ii) The number of cards in the top row (7th row) is given by a7.
a7=14+7-1-2                            an=a+n-1d=14-12=2

Hence, the correct answer is option (d).

(iii) The mathematical concept applied in solving the above problem is Arithmetic Progression.

Hence, the correct answer is option (c).
 



Page No 81:

Question 54:

Each of the following questions contains STATEMENT-1 (Assertion) and STATEMENT-2 (Reason) and has following four choices (a), (b), (c) and (d), only one of which is the correct answer. Mark the correct choice.
(a) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.
(b) Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.
(c) Statement-1 is true, Statement-2 is false.
(d) Statement-1 is false, Statement-2 is true.
Statement-1 (Assertion): The nth term an, of an A.P., the sum of whose n terms is Sn, is given by an = SnSn – 1, n > 1.
Statement-2 (Reason): The common difference ‘d’ of an A.P., the sum of whose n terms Sn is given by d = Sn – 2Sn – 1 + Sn – 2, n > 2.

Answer:

Statement-1 (Assertion): The nth term an, of an A.P., the sum of whose n terms is Sn, is given by an = Sn – Sn – 1n > 1.
Sn=n22a+n-1dSn-1=n-122a+n-2dSn-Sn-1=n22a+n-1d-n-122a+n-2dSn-Sn-1=n22a+n-1d-n22a+n-2d+122a+n-2dSn-Sn-1=nd2+a+nd2-dSn-Sn-1=a-d+ndSn-Sn-1=a+dn-1=an

Thus, Assertion is true.

Statement-2 (Reason): The common difference ‘d’ of an A.P., the sum of whose n terms Sn is given by d = Sn – 2Sn – 1 + Sn – 2n > 2.
Sn=n22a+n-1dSn-1=n-122a+n-2dSn-2=n-222a+n-3dSn-2Sn-1+Sn-2=n22a+n-1d-2n-122a+n-2d+n-222a+n-3dSn-2Sn-1+Sn-2=n22a+dn-d-2n-222a+dn-2d+n-222a+dn-3dSn-2Sn-1+Sn-2=2an2+n2d2-nd2-4an-4a+2n2d-4nd-2nd+4d2+2an-4a+n2d-2dn-3dn+6d2Sn-2Sn-1+Sn-2=2an2+n2d2-nd2-2an+2a-n2d+2nd+nd-2d+an-2a-nd+3d+n2d-3dn2Sn-2Sn-1+Sn-2=an-2an+an+n2d2+n2d2-n2d+-nd2+nd-nd-3nd2+2nd+2a-2a+-2d+3dSn-2Sn-1+Sn-2=d

Thus, Statement-2 is also true.

But Statement-2 is not a correct explanation for Statement-1.

Hence, the correct answer is option (b).

Page No 81:

Question 55:

Each of the following questions contains STATEMENT-1 (Assertion) and STATEMENT-2 (Reason) and has following four choices (a), (b), (c) and (d), only one of which is the correct answer. Mark the correct choice.
(a) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.
(b) Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.
(c) Statement-1 is true, Statement-2 is false.
(d) Statement-1 is false, Statement-2 is true.
Statement-1 (Assertion): The sum of n terms of the series 5+20+45+80+ ... is 52nn+1.
Statement-2 (Reason): The sum of first n natural numbers is nn+12.

Answer:

Statement-1 (Assertion): The sum of n terms of the series 5+20+45+80+ ... is 52nn+1.

5+20+45+80+ ...n terms=5+25+35+45+ ...n terms=51+2+3+4+...n terms=5nn+12                              Sum of first n natural numbers,Sn=n2n+1=52nn+1

Thus, Statement-1 is true.

Statement-2 (Reason): The sum of first n natural numbers is nn+12.
The sum of first n positive integers is given as: 1 + 2 + 3 + 4 + ...
Sn=n21+n                             Sn=n2a+l

Thus, Statement-2 is true and it is the correct explanation of Statement-1.

Hence, the correct answer is option (a).

Page No 81:

Question 56:

Each of the following questions contains STATEMENT-1 (Assertion) and STATEMENT-2 (Reason) and has following four choices (a), (b), (c) and (d), only one of which is the correct answer. Mark the correct choice.
(a) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.
(b) Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.
(c) Statement-1 is true, Statement-2 is false.
(d) Statement-1 is false, Statement-2 is true.
Statement-1 (Assertion): The sum of n terms of an AP with first and last terms as a1 and an respectively, is Sn=n2a1+an.
Statement-2 (Reason): The sum of the terms equidistant from the beginning and end in the A.P. a1, a2, a3, ..., an – 2, an – 1, an is equal to a1 + an.

Answer:

Statement-2 (Reason): The sum of the terms equidistant from the beginning and end in the A.P. a1a2a3, ..., an – 2an – 1an is equal to a1 + an.

Let the common difference of the AP be d. Then, kth term from the beginning is ak and kth term from last is the (nk + 1)th term from beginning i.e., an-k+1.
Thus,
 ak = a1 + (– 1)d                             [∵  an = a1 + (– 1)d ]
And,
an-k+1=a1+n-k+1-1d              an=a1+n-1d=a1+n-kd
Now,
ak+an-k+1=a1+k-1d+a1+n-kd=2a1+dn-1=a1+a1+dn-1=a1+an

Thus, Statement-2 is true.

Statement-1 (Assertion): The sum of n terms of an AP with first and last terms as a1 and an respectively, is Sn=n2a1+an.
The sum of the first n terms of an AP with first term a and common difference d is given as: Sn=n22a+n-1d.
Sn=a1+a2+a3+a4+...+an-3+an-2+an-1+anSn=a1+an+a2+an-1+a3+an-2+a4+an-3+...          
Sn=a1+an +a1+an +a1+an +a1+an +...n2times       From statement2Sn=n2a1+an     

Thus, Statement-1 is true.
And Statement-2 is the correct explanation of Statement-1.

Hence, the correct answer is option (a).

Page No 81:

Question 57:

Each of the following questions contains STATEMENT-1 (Assertion) and STATEMENT-2 (Reason) and has following four choices (a), (b), (c) and (d), only one of which is the correct answer. Mark the correct choice.
(a) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.
(b) Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.
(c) Statement-1 is true, Statement-2 is false.
(d) Statement-1 is false, Statement-2 is true.
Statement-1 (Assertion): The sum of first n even natural numbers is n(n + 1).
Statement-2 (Reason): The sum of first n odd natural numbers is n(n – 1).

Answer:

Statement-1 (Assertion): The sum of first n even natural numbers is n(n + 1).

Consider the sum of first n natural numbers, i.e., Sn=1+2+3+4+...
Therefore, Sum of first n natural numbers,Sn=n2n+1.                  .....(1)

Now, consider the sum of first n even numbers (nE), i.e., SnE=2+4+6+8+...
Sn2=2+4+6+8+...=21+2+3+4+...=2×nn+12                             Using 1=nn+1

Thus, Statement-1 is true.

Statement-2 (Reason): The sum of first n odd natural numbers is n(n – 1).
Now, consider the sum of first n odd numbers (nO), i.e., SnO=1+3+5+7+...
Sn3=1+3+5+7+...=n22×1+n-12                          Sn=n22a+n-1d=n22+2n-2=n22n=n2

Thus, Statement-2 is false.

Hence, the correct answer is option (c).

Page No 81:

Question 58:

Each of the following questions contains STATEMENT-1 (Assertion) and STATEMENT-2 (Reason) and has following four choices (a), (b), (c) and (d), only one of which is the correct answer. Mark the correct choice.
(a) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.
(b) Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.
(c) Statement-1 is true, Statement-2 is false.
(d) Statement-1 is false, Statement-2 is true.
Statement-1 (Assertion): If a1, a2, a3, ..., an is an AP such that a1 + a4 + a7 + … + a16 = 147, then a1 + a6 + a11 + a16 = 98.
Statement-2 (Reason): In an A.P., the sum of the terms equidistant from the beginning and the end is always same and is equal to the sum of first and last term.

Answer:

Statement-2 (Reason): In an A.P., the sum of the terms equidistant from the beginning and the end is always same and is equal to the sum of first and last term.

Let the common difference of the AP be d. Then, kth term is ak = a1 + (– 1)d and kth term from last is the (n – k + 1)th term from beginning.
Thus,
an-k+1=a1+n-k+1-1d              an=a1+n-1d=a1+n-kd
Now,
ak+an-k+1=a1+k-1d+a1+n-kd=2a1+dn-1=a1+a1+dn-1=a1+an

Thus, Statement-2 is true. 


Statement-1 (Assertion): If a1a2a3, ..., an is an AP such that a1 + a4 + a7 + … + a16 = 147, then a1 + a6 + a11 + a16 = 98.

Given that, a1a2a3, ..., an is an AP such that a1 + a4 + a7 + … + a16 = 147. Let the common difference of the AP be d.
a1+a4+a7++a16=147a1+a1+3d+a1+6d++a1+15d=147                    an=a1+n-1d6a1+3d1+2+...+5=1476a1+3d×5×62=1476a1+45d=1472a1+15d=49a1+a1+15d=49a1+a16=49

Now,
a1+a6+a11+a16=a1+a1+5d+a1+10d+a1+15d=4a1+30d=22a1+15d=2a1+a1+15d=2a1+a16=2×49=98

Thus, Statement-1 is true.
And Statement-2 is the correct explanation of Statement-1.

Hence, the correct answer is option (a).



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