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If the sum of $n$ terms of an A.P. is $3 n^2+5 n$ and its $m^{\text {th }}$ term is 164 , find the value of $m$.
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Let the sum of $n$ term is denoted by $S_n$
$\therefore S_n=3 n^2+5 n$
Put $n=1,2$. $T_1=S_1=3.1^2+5.1=3+5=8$;
$S_2=T_1+T_2=3 \cdot 2^2+5.2=12+10=22$
$\therefore \quad T_2=S_2-S_1=22-8=14$
$\therefore$ Common difference $d=T_2-T_1=14-8=6$
$a=8, \quad d=6$
$m^{\text {th }}$ term $=a+(m-1) d=164 \Rightarrow 8+(m-1) \cdot 6=164$
$6 m+2=164 \Rightarrow 6 m=164-2=162$
$\therefore \quad m=\frac{162}{6}=27$
$\therefore S_n=3 n^2+5 n$
Put $n=1,2$. $T_1=S_1=3.1^2+5.1=3+5=8$;
$S_2=T_1+T_2=3 \cdot 2^2+5.2=12+10=22$
$\therefore \quad T_2=S_2-S_1=22-8=14$
$\therefore$ Common difference $d=T_2-T_1=14-8=6$
$a=8, \quad d=6$
$m^{\text {th }}$ term $=a+(m-1) d=164 \Rightarrow 8+(m-1) \cdot 6=164$
$6 m+2=164 \Rightarrow 6 m=164-2=162$
$\therefore \quad m=\frac{162}{6}=27$
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