Question

Which term of the AP $3,15,27,39,...$ will be $120$ more than its ${21}^{st}$ term ?

Answer 1

A.P $=3,15,27,39,..........$

From the above A.P , we know that

$a=3$

and $d=12$

$A_{21}=a+20d$

$=3+240$

$=243$

Lets suppose the term of AP will be 120 more than its ${21}^{st}$ term be $A_n$

$A_n=A_{21}+120=243+120=363$

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

$363=3+(n-1)12$

$363-3=(n-1)12$

$\frac{360}{12}=n-1$

$n-1=30$

$n=31$

so, ${31}^{st}$ term of the AP will be $120$ more than its ${21}^{st}$ term.