Question

Find four consecutive terms in an A.P. whose sum is $-54$ and the sum of $1^{th}$ and $3^{rd}$ term is $-30.$

• A. $-18,-15,-12,-9$
• B. $-18,-12,-6,0$
• C. $-20,-15,-10,-5$
• D. $-20,-18,-16,-14$

Answer 1

The correct option is A $-18,-15,-12,-9$

Let the terms be $(a-3d),(a-d),(a+d),(a+3d).$ Then,

$(a-3d)+(a-d)+(a+d)+(a+3d)=-54$

$\Rightarrow 4a=-54\Rightarrow a=\frac{-27}{2}$

Also, Sum of $1^{st}$ and $3^{rd}=-30$

$\Rightarrow (a-3d)+(a+d)=-30\Rightarrow 2a-2d=-30$

$\Rightarrow a-d=-15\Rightarrow d=a+15=\frac{-27}{2}+15=\frac{3}{2}$

Therefore, the four terms are $(\frac{-27}{2}-3\times \frac{3}{2}),(\frac{-27}{2}-\frac{3}{2}),(\frac{-27}{2}+\frac{3}{2}),(\frac{-27}{2}+3\times \frac{3}{2})$
i.e.$-18,-15,-12,-9$