Question

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

Answer 1

Let the four consecutive terms in an A.P. be a – 3d, a – d, a + d and a + 3d.
We know that the sum of these four consecutive numbers is –54.
Thus, we have:
a – 3d + a – d + a + d + a + 3d = –54
$\Rightarrow$4a = –54
$\Rightarrow$a =$\frac{-54}{4}=\frac{-27}{2}$ ...(i)
Also,
The sum of the 1^st and 3^rd terms is –30.
Thus, we have:
a – 3d + a + d = –30
$\Rightarrow$2a – 2d = –30
$\Rightarrow$2 ×$\left(\frac{-27}{2}\right)$ – 2d = –30
$\Rightarrow$–27 – 2d = –30
$\Rightarrow$–2d = –30 + 27 = –3
$\Rightarrow$d = $\frac{-3}{-2}=\frac{3}{2}$
Now,
$$\begin{gathered} a-3d=\frac{-27}{2}-3\times \frac{3}{2}=\frac{-27}{2}-\frac{9}{2}=\frac{-36}{2}=-18 \\ a-d=\frac{-27}{2}-\frac{3}{2}=\frac{-30}{2}=-15 \\ a+d=\frac{-27}{2}+\frac{3}{2}=\frac{-24}{2}=-12 \\ a+3d=\frac{-27}{2}+3\times \frac{3}{2}=\frac{-27}{2}+\frac{9}{2}=\frac{-18}{2}=-9 \end{gathered}$$

Therefore, the four consecutive numbers in the A.P. are –18, –15, –12 and –9.