Question

An A.P. consists of $13$ terms of which $2^{nd}$ terms is $10$ and the last term is $120$. Find the $9^{th}$ term.

• A. $70$
• B. $80$
• C. $90$
• D. $100$

Answer 1

Answer: (B)

$80$

Given that, $a_2=10,a_{13}=120,n=13$
We know that, $a_n=a+(n-1)d$
$\Rightarrow a_2=a+(2-1)d$
$\Rightarrow 10=a+d$ ....... (1)
Similarly, $a_{13}=a+(13-1)d$
$\Rightarrow 120=a+12d$ ...... (2)
On subtracting equation (1) & (2), we get
$-110=-11d$
$\Rightarrow d=10$
Put the value of $d$ in equation (1), we get
$10=a+10$
$\Rightarrow 10-10=a$
$\Rightarrow a=0$
We need to find $a_9=a+(9-1)d$
$=0+8\left(10\right)$
$=0+80$
$=80$
Hence, $9^{th}$ term is $80$.