Question

Find four consecutive terms of an A.P. whose sum is $88$ and sum of first and third term is $40$.

Answer 1

Let four consecutive terms are a - 3d , a - d , a + d , a + 3d , where a and d is real numbers.

According to question,

sum of all four terms = 88

$(a-3d)+(a-d)+(a+d)+(a+3d)=88$

$4a=88$

$a=22$ ......(1)

sum of 1st and 3rd term is 40

$(a-3d)+(a+d)=40$

$2a-2d=40$

$a-d=20$

from equation (1),

$22-d=20$

$d=2$

hence, $a=22$ and $d=2$

then, $a-3d=22-3\times 2=22-6=16$

$a-d=22-2=20$

$a+d=22+2=24$

$a+3d=22+3\times 2=22+6=28$

therefore, four consecutive term in an A.P are $16,20,24,28$