Question

The sum of first m terms of an A.P. is $4m^2$-m. If its nth terms is 107, find the value of n. Also, find the 21st term of this A.P.

Answer 1

we are given sum of First m terms

$S_m=4m^2-m$

Substitute different values of m starting from 1,2,3....

$S_1=4(1{)}^2-(1)=3$

$S_2=4(2{)}^2-(2)=14$

$S_3=4(3{)}^2-(3)=33$

$S_1=a_1=3$

$S_2-S_1=a_2=14-3=11$

$S_3-S_2=a_3=33-14=19$

Therefore, A.P is 3,11,19,....

a= 3, d= 8

Now nth term is 107

$a_n=107$

$a+(n-1)d=107$

$a+(n-1)8=107$

$a+8n-8=107$

$3+8n-8=107$

$8n=112$

$n=\frac{112}{8}=14$

Now calculate 21st term as

$a_{21}=a+20d$

$a_{21}=3+20\times 8$

$a_{21}=163$