If S7=49 and S17=289 then find …

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Yogita Ingle 6 years, 3 months ago
Formula of first n terms in AP = {tex}S_n = \frac{n}{2}(2a+(n-1)d){/tex}
Substitute n = 7
So,{tex} S_7 = \frac{7}{2}(2a+6d){/tex}
{tex}49 = \frac{7}{2}(2a+6d){/tex}
{tex}49 \times \frac{2}{7} = 2a+6d{/tex}
14= 2a+6d
Substitute n = 17
{tex}S_{17} = \frac{17}{2}(2a+16d){/tex}
{tex}289 = \frac{17}{2}(2a+16d){/tex}
{tex}289 \times \frac{2}{17} = 2a+16d{/tex}
34= 2a+16d
34= 2a+10d+6d
Using A
34= 14+10d
20 = 10d
d=2
Substitute the value of d in A
14= 2a+12
2= 2a
a=1
So, {tex} S_n = \frac{n}{2}(2(1)+(n-1)2){/tex}
{tex}S_n = \frac{n}{2}(2+(n-1)2){/tex}
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