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.
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 $1,6,11,16,.......$
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