Question

Find four consecutive terms in an A.P whose sum is $12$ and sum of $3^{rd}$ and $4^{rd}$ term is $14$.
(Assume the four consecutive terms in A.P. are $a-d,a,a+d,a+2d$)

Answer 1

Let the terms be $a-d,a,a+d,a+2d$

Sum$=4a+2d=12$

$2a+d=6$------------(1)

Sum of 3rd & 4th term$=(a+d)+(a+2d)=14$

$2a+3d=14$-----------(2)

By substracting Equation (1) from (2)

$-2d=-8$

$\Rightarrow d=4$

From Equation(1), $a=1$

So, the terms are $(1-4),1,(1+4),(1+2(4\left)\right)$

$=-3,1,5,9$