If the sum of 7 terms of an A.P. is 49 and that of 17 terms is 289, find the sum of n terms.
Given:
The sum of 7 terms of an A.P. is 49 and that of 17 terms is 289.
To do:
We have to find the sum of $n$ 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)$
$S_{7}=\frac{7}{2}[2(a)+(7-1)d]$
$49=\frac{7}{2}(2a+6d)$
$49=7(a+3d)$
$a+3d=7$
$a=7-3d$......(i)
$S_{17}=\frac{17}{2}[2(a)+(17-1)d]$
$289=\frac{17}{2}(2a+16d)$
$289=17(a+8d)$
$a+8d=17$
$7-3d+8d=17$ (From (i))
$5d=17-7$
$d=\frac{10}{5}$
$d=2$
This implies,
$a=7-3(2)$
$a=7-6$
$a=1$
The sum of $n$ terms $S_n=\frac{n}{2}(2a+(n-1)d)$
$S_n=\frac{n}{2}[2(1)+(n-1)2]$
$=n(1+n-1)$
$=n^2$
Hence, the sum of $n$ terms is $n^2$.
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