Question

Find four consecutive terms in an A.P. whose sum is 12 and sum of 3^rd and 4^th term is 14.

(Assume the four consecutive terms in A.P. are a – d, a, a + d, a +2d)

Answer 1

Assume that the four consecutive terms in A.P. are a – d, a, a + d, a +2d .

It is given that,
Sum of four consecutive terms = 12
Sum of 3^rd and 4^th term = 14

$$\begin{gathered} \left(a-d\right)+a+\left(a+d\right)+\left(a+2d\right)=12 \\ \Rightarrow 4a+2d=12 \\ \Rightarrow 2a+d=6 \\ \Rightarrow 2a=6-d...\left(1\right) \end{gathered}$$

$$\begin{gathered} \left(a+d\right)+\left(a+2d\right)=14 \\ \Rightarrow 2a+3d=14 \\ \Rightarrow 6-d+3d=14 \\ \Rightarrow 2d=14-6 \\ \Rightarrow 2d=8 \\ \Rightarrow d=4\left(\mathrm{from}\left(1\right)\right) \\ \Rightarrow 2a=6-d \\ \Rightarrow 2a=6-4 \\ \Rightarrow 2a=2 \\ \Rightarrow a=1 \end{gathered}$$

Hence, the terms are –3, 1, 5 and 9.