Question

The sum of first $16$ terms of an $A.P$ is $112$ and sum of its next fourteen terms is $518$. Find the $A.P$.

Answer 1

Given $S_{16}=112$

sum of next 14 terms = 518

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

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

$112=8[2a+15d]$

$14=2a+15d\cdots \left(1\right)$

sum of next 14 terms is 518

sum of (16 + 14) terms $\left(S_{30}\right)=112+518$

$S_{30}=630$

$S_{30}=\frac{30}{2}[2a+(30-1\left)d\right]$

$630=15[2a+29d]$

$42=2a+29d\cdots \left(2\right)$

Subtracting equation (1) from (2)

$42=2a+29d$

$14=2a+15d$

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_____________

$28=14d.$

$d=28/14$

$d=2.$

Substitute value of d in equation (1)

$14=2a+15d$

$14=2a+15\left(2\right)$

$14=2a+30$

$2a=-16$

$a=-16/2$

$a=-8$

If 'a' is the first term and 'd' is a common difference then the A.P is

$a,a+d,a+2d,a+3d\cdots$

$-8,-6,-4,-2\cdots$