Question

the sum of four consecutive numbers in an AP is 32 and the ratio of the product of the first and the last terms to the product of the two middle terms is 7:15.find the nos? plz its urgent.

Answer 1

Let the four consecutive terms in AP be a-3d,a-d,a+d,a+3d.
a-3d+a-d+a+d+a+3d=32
4a=32
a=8
Four numbers = 8-3d,8-d,8+d and 8+3d.
8-3d*8+3d:8-d*8+d=7:15
64-9d²:64-d²=7:15
64-9d²*(15)=64-d²*(7)
960-135d²=448-7d²
512=128d²
d²=4
d=2
Therefore, four numbers = 8-3*2,8-2,8+2,8+3*2
Four numbers = 8-6,6,10,8+6
Four numbers = 2,6,10,14

Answer 2

let a is the first term of AP and d is the common difference
let four consecutive term ,
a-3d, a-d, a+d, a+3d
now ,
a/c to question ,
a-3d+a-d+a+d+a+3d=32
4a=32
a=8
now again question ask
( a-3d)(a+3d)/(a-d)(a+d)=7/15

=>15a^2-135d^2=7a^2-7d^2

=> 8a^2=128d^2

=> 8 (8)^2=64 x 2d^2

=> d=+_2

hence four numbers:-
2,6,10,14 or, 14,10,6,2