Question
Determine the $A.P.$ whose third term is $5$ and seventh term is $9$ .

Answer

Given, $T_3=5$ and $T_7=9$
We know that, $n^{\text {th }}$ term of an $A.P.$ is
$T_n=a+(n-1) d$
$5=a+4 d$
$\therefore 9=a+6 d$ and $9=a+6 d$
$Eq(i)-Eq \text { (ii), }$
$-4=-2 d $
$\Rightarrow d=2$
From eq $(i),$
$5=a+4(2)$
$\Rightarrow a=5-8=-3$
So, required $A.P.$ is $-3,-1,1,3,5,7,9, \ldots$.

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