Question

4. An AP consists of 63 terms. If its 32nd term is q, then sum of all the terms of this
AP is
(a) 63q
(b) 32q
(c) 31q
(d) 95q​

Answer 1

32nd term = q

a+31d= q

thus,

a= q-31d......

n= no. of terms

thus,

Sn= n/2( 2a+ (n-1)d)

Sn= 63/2( 2q-62 d + 62d)

thus, answer is 63q

Answer 2

Given:

Let the first term = a and common difference = d

$a_{32}$ = q

We have to find, the value of sum of all the terms of this  AP is:

Solution:

We know that,

The nth term of an AP

$a_{n}$ = a +(n - 1)d

∴ The 32nd term of an AP

$a_{32}$ = a + (32 - 1)d

$a_{32}$ = a + 31d              .......... (1)

The sum of an AP

$S_{n} =\dfrac{n}{2} [2a+(n-1)d]$

∴ $S_{63.} =\dfrac{63}{2} [2a+(63-1)d]$

⇒ $S_{63.} =\dfrac{63}{2} [2a+(63-1)d]$

⇒ $S_{63.} =\dfrac{63}{2} [2a+62d]$

⇒ $S_{63.} =\dfrac{63}{2} [2(a+31d)]$

⇒ $S_{63}$ = 63(a + 31d)

Using equation (1), we get

$S_{63}$ = 63q

∴ The sum of all the terms of this  AP , $S_{63}$ = 63q

Thus, the required "option a) 63q is correct".