Question

If $a_1,a_2,a_3,a_4,..{..,a}_{20}$ are in AP and $a_1+a_{20}=45$, then $a_1+a_2+a_3+a_4,..{..+a}_{20}$ is equal to

• A. $90$
• B. $350$
• C. $900$
• D. $730$
• E. $450$

Answer 1

Answer: (E)

The correct option is D $450$
$\because$ Given $a_1,a_2,a_3,....a_{20}$ are in AP and
$a_1+a_{20}=\mathrm{45..}\left(i\right)$
sum of first $n$ terms$=\frac{n}{2}(a+l)$
$a+l=45$
$[\because a_1+a_{20}=45.]$
Have, $n=20$
$\therefore a_1+a_2+a_3+....{+a}_{20}=\frac{20}{2}\times 45=450$