# The sum of the $5^{\text {th }}$ and the $7^{\text {th }}$ terms of an AP is 52 and the $10^{\text {th }}$ term is 46 . Find the AP.

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Given:

The sum of the $5^{\text {th }}$ and the $7^{\text {th }}$ terms of an AP is 52 and the $10^{\text {th }}$ term is 46.

To do:

We have to find the AP.

Solution:

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

This implies,

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

$46=a+9d$

$a=46-9d$.........(i)

$a_5=a+(5-1)d$

$=a+4d$

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

$=a+6d$

Therefore,

$a_5+a_7=52$

$a+4d+a+6d=52$

$2a+10d=52$

$2(46-9d)+10d=52$        [From (i)]

$92-18d+10d=52$

$92-52=8d$

$d=\frac{40}{8}$

$d=5$

This implies,

$a=46-9(5)$

$=46-45$

$=1$

This implies,

$a_2=a+d=1+5=6$

$a_3=a+2d=1+2(5)=1+10=11$

$a_4=a+3d=1+3(5)=1+15=16$

The required AP is $1,6,11,16,.......$

Updated on 10-Oct-2022 13:27:34