Question

plz solve ques 12 and 13

Answer 1

12) Let the terms be (a-3d), (a-d), (a+d) and (a+3d)
here the first term is (a-3d) and common difference is 2d

A/Q
(a-3d) + (a-d) + (a+d) + (a+3d) = 32
=> 4a = 32
=> a = 8

(a-3d)*(a+3d) / (a-d)*(a+d) = 7/15
=> (a^2 - 9d^2)/ (a^2 - d^2) = 7/15
=> 15(a^2 - 9d^2) = 7(a^2 - d^2)
=> 15a^2 - 135d^2 = 7a^2 - 7d^2
=> 8a^2 = 128d^2
=> a^2 = 16d^2
Substituting value of a as 8
=> 64 = 16d^2
=>4 = d^2
=> d=2 or d=-2

Hence the numbers are 2, 6, 10, 14


13)
Let the numbers be (a-d) , a , (a+d)

A/Q
(a-d) + a + (a+d) = 48
=> 3a = 48
=> a= 16

(a-d)*a = 4(a+d) +12
=>(16-d)*16 = 4(16+d)+12
=>256-16d=64+4d+12
=>180=20d
=>d=9

Hence numbers are 7, 16, 25