Question

Sum of first 10 terms of an ap is 100 and first 25 terms is 625. Find the number of terms in it,whose sum is 441.

Answer 1

Please see the attachment

Answer 2

Given:

Sum of first 10 terms of an ap is 100 and first 25 terms is 625.

To Find :

Find the number of terms in it,whose sum is 441.

Solution:

Formula of sum of first n terms :

$S_n=\frac{n}{2}(2a+(n-1)d)$

Substitute n = 10

$S_{10}=\frac{10}{2}(2a+(10-1)d)\\S_{10}=5(2a+9d)\\S_{10}=10a+45d$

We are given that Sum of first 10 terms of an ap is 100

So,10a+45d=100

2a+9d=20---1

Substitute n = 25

$S_{25}=\frac{25}{2}(2a+(25-1)d)\\S_{25}=\frac{25}{2}(2a+24d)$

We are given that Sum of first 25 terms of an ap is 625

So, $625=\frac{25}{2}(2a+24d)\\625 \times 2 =25(2a+24d)$

2a+24d=50 ---2

Substitute the value of 2a from 2 in 1

50-24d+9d=20

50-15d=20

30=15d

d=2

Substitute the value of d in 1

2a+9d=20

2a+9(2)=20

2a=2

a=1

We are supposed to find  the number of terms in it,whose sum is 441.

So, $S_n=\frac{n}{2}(2a+(n-1)d)\\441=\frac{n}{2}(2(1)+(n-1)(2))\\n=-21,21$

Since no. of terms cannot be negative

So, The number of terms in it whose sum is 441 is 21