Question

Which terms of the A.P.3,15,27,39,... will be 120 more than its 21st term ?

Answer 1

The given A.P.3,15,27,39,

Here First term a =3

Common difference d= 12

Hence $T_{21}$ = a +(21-1)d

=a+20d

=3+240

=243

Also given

$T_n$ = 120 + 243 Further we know that $T_n$ = a+ (n-1) d
= 363 = 3 + (n-1) d
Equating 363= $T_n$ = 3 + (n-1) d
363=3+(n-1)12

or 363= 3 +12n -12

or 363 = 12n -9

or 363+9 =12n

or 372 =12n

$\frac{372}{12}$ =n

n=31
Hence the 31st term = 363