Question

In a AP given that S20=S40 then S60=?


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Answer 1

Given:

In an AP, sum of first 20 terms = sum of first 40 terms

To find:

Sum of first 60 terms.

Solution:

The sum of $n$ terms in an AP with its first term being $a$ and the  common difference $d$ is given by:

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

where, $S_{n}$ is the sum of $n$ terms.

Here, we are given that the sum of the first 20 terms and first 40 terms are equal.

$S_{20}=\frac{20}{2}[2a+(20-1)d]$

$S_{40}=\frac{40}{2}[2a+(40-1)d]$

Now, $S_{20}=S_{40}$

$\frac{20}{2}(2a+19d) =\frac{40}{2}(2a+39d)$

$2a+19d=2(2a+39d)$

$2a+19d=4a+78d$

$19d-78d=4a-2a$

$-59d=2a$

$2a+59d=0...(1)$

We have to find the sum of the first 60 terms, which is given by:

$S_{60}=\frac{60}{2}[2a+(60-1)d]$

$S_{60}=15(2a+59d)...(2)$

From equation (1), we have

$2a+59d=0$

Substituting equation (1) in equation (2),

$S_{60}=15(0)=0$

Hence, the sum of the first 60 terms in the AP is 0.

If the sum of the first 20 terms is equal to the sum of the first 40 terms, then the sum of the first 60 terms in the AP is 0.