If sum of the $ 3^{\text {rd }} $ and the $ 8^{\text {th }} $ terms of an AP is 7 and the sum of the $ 7^{\text {th }} $ and the $ 14^{\text {th }} $ terms is $ -3 $, find the $ 10^{\text {th }} $ term.

Given: 

The sum of the \( 3^{\text {rd }} \) and the \( 8^{\text {th }} \) terms of an AP is 7 and the sum of the \( 7^{\text {th }} \) and the \( 14^{\text {th }} \) terms is \( -3 \).

To do: 

We have to find the \( 10^{\text {th }} \) term.

Solution:

Let $a$ be the first term and $d$ be the common difference.

This implies,

$a_{3}=a+(3-1)d$

$=a+2d$.......(i)

$a_8=a+(8-1)d$

$=a+7d$.........(ii)

$a_7=a+(7-1)d$

$=a+6d$.........(iii)

$a_{14}=a+(14-1)d$

$=a+13d$........(iv)

According to the question,

$a_3+a_8=7$

$a+2d+a+7d=7$        [From (i) and (ii)]

$2a+9d=7$.......(v)

$a_7+a_{14}=-3$

$a+6d+a+13d=-3$

$2a+19d=-3$.......(vi)

Subtracting (v) from (vi), we get,

$2a+19d-(2a+9d)=-3-7$

$10d=-10$

$d=-1$

This implies,

$2a+9(-1)=7$

$2a=7+9$

$2a=16$

$a=8$

Therefore,

$a_{10}=a+(10-1)d$

$=8+9(-1)$

$=8-9$

$=-1$

Hence, the \( 10^{\text {th }} \) term of the given AP is $-1$.

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Updated on: 10-Oct-2022

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