Class 11

Math

Algebra

Sequences and Series

If the sum of n terms of an A.P. is $3n_{2}+5n$ and its mth term is 164, find the value of m.

$2n (2a+((n−1)d)=3n_{2}+5n$

$(2a−d)+nd=6n+10$

$d=6and2a−d=10$

$d=6anda=8$

Now, $m_{th}$ term $=a+(m−1)d=164$

$⇒8+6(m−1)=164$

$⇒6(m−1)=156$

$⇒m=27$