The sum of first 6 terms of an $ \mathrm{AP} $ is 36 and the sum of its first 16 terms is 256. Find the sum of first 10 terms of this $ A P $.


Given:

The sum of first 6 terms of an \( \mathrm{AP} \) is 36 and the sum of its first 16 terms is 256.

To do:

We have to find the sum of its first $10$ terms.

Solution:

Let the first term be $a$ and the common differnce be $d$.

We know that,

Sum of $n$ terms$ S_{n} =\frac{n}{2}(2a+(n-1)d)$

$S_{6}=\frac{6}{2}[2(a)+(6-1)d]$

$36=3(2a+5d)$

$12=2a+5d$

$2a=12-5d$......(i)

$S_{16}=\frac{16}{2}[2(a)+(16-1)d]$

$256=8(2a+15d)$

$32=2a+15d$

$12-5d+15d=32$       (From (i))

$10d=32-12$

$d=\frac{20}{10}$

$d=2$

This implies,

$2a=12-5(2)$

$2a=12-10$

$a=\frac{2}{2}$

$a=1$

The sum of $10$ terms $S_{10}=\frac{10}{2}[2(1)+(10-1)2]$

$=5[2+9(2)]$

$=5(2+18)$

$=5(20)$

$=100$

Hence, the sum of $10$ terms is $100$.   

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Updated on: 10-Oct-2022

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