Question

The fourth term of an A.P. is $11$ and the eighth term exceeds twice the fourth term by $5$. Find the A.P. and the sum of first $50$ terms.

• A. $3540$
• B. $6754$
• C. $4850$
• D. $7819$

Answer 1

The correct option is C $4850$

$\Rightarrow$$4^{th}term=11$

$\therefore$ $a+3d=11$

$\Rightarrow$$8^{th}term=2\left(4^{th}term\right)+5$

$\Rightarrow$$a+7d=27$

$\Rightarrow$$(a+7d)-(a-3d)=27-11$

$\Rightarrow$$4d=16$

$\Rightarrow$$d=4$

$\Rightarrow$$a=-1$

Formula: $S_n=\frac{n}{2}[2a+(n-1\left)d\right]$

Sum of first $50$ terms $=\frac{50}{2}\left(2\right(-1)+49(4\left)\right)$
$=25\left(194\right)$

$=4850$