Question

find 4 consecutive terms in an A.P. whose sum is 36 & the product of 2nd and the 4th is 105. the terms are in ascending order .​

Answer 1

$\huge\mathfrak\red{☟ \: \: \: answer \: \: \: \: ✎}$

let the 4 terms be a - 3d, a-d, a+d and a+3d

Sum = 4a = 36, so a = 9

Product of 2nd and 4th = (9-d)(9+3d)=105

81+27d-9d-3d2 = 105

3d2-18d+24=0

d2-6d+8=0

(d-2)(d-4)=0

so d=2, or 4

numbers are

3, 7, 11, 15

or

-3, 5, 13, 21

$\\ \\ \\ \sf \colorbox{gold} {\red(ANSWER ᵇʸ ⁿᵃʷᵃᵇ⁰⁰⁰⁸}$