Question

Which term of the arithmetic progression $5,15,25,...$ will be $130$ than its ${31}^{st}$ term?

Answer 1

We have, $a=5$ and $d=10$

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

$\therefore a_{31}=a+30d=5+30\times 10=305$

Let $n^{th}$ term of the given A.P. be 130 more than its ${31}^{st}$ term.

Then, $a_n=130+a_{31}$

$\therefore a+(n-1)d=130+305$

$⟹5+10(n-1)=435$

$⟹10(n-1)=430$

$⟹n-1=43$

$⟹n=44$

Hence, ${44}^{th}$ term of the given A.P. is $130$ more than its ${31}^{st}$ term.