Class 11 Math Algebra Sequences and series

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

Solution:

Given $S_{16}=112$

sum of next 14 terms = 518

$S_{n}=2n [2a+(n−1)d]$

$S_{16}=216 [2a+(16−1)d]$

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

$14=2a+15d⋯(1)$

sum of next 14 terms is 518

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

$S_{30}=630$

$S_{30}=230 [2a+(30−1)d]$

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

$42=2a+29d⋯(2)$

Subtracting equation (1) from (2)

$42=2a+29d$

$14=2a+15d$

- - -

_____________

$28=14d.$

$d=28/14$

$d=2.$

Substitute value of d in equation (1)

$14=2a+15d$

$14=2a+15(2)$

$14=2a+30$

$2a=−16$

$a=−16/2$

$a=−8$

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

$a,a+d,a+2d,a+3d⋯$

$−8,−6,−4,−2⋯$

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