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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