Question

Find the ${21}^{st}$ term of an A.P. whose $1^{st}$ term is $8$ and the ${15}^{th}$ term is $120$.

• A. $167$
• B. $165$
• C. $168$
• D. $169$

Answer 1

The correct option is C $168$

Given that, $a_1=8;a_{15}=120$
We know that, $a_n=a+(n-1)d$
$\Rightarrow a_1=a+(1-1)d$
$\Rightarrow 8=a+\left(0\right)d$ ....... (1)
Similarly, $a_{15}=a+(15-1)d$
$120=a+14d$ ..... (2)
On subtracting equation (1) & (2), we get
$112=14d$
$\Rightarrow d=8$
Now we have to find $a_{21}$

Thus $a_{21}=a+(21-1)d$
$=8+20\left(8\right)$
$=8+160$
$=168$
Hence, ${21}^{st}$ term is $168$.