Question

The sum of the foure
a consecutive terms which
are in an arithmetic progression is 32 and the ratio of the product of first and last term to the product of 2 middle term is 7 isto 15 find the number
​

Answer 1

Step-by-step explanation:

let the terms are (a-3d),(a-d),(a+d),(a+3d)

where a is the 1st term of the series and 2d is the common difference.

so,

(a-3d)+(a-d)+(a+d)+(a+3d)=32

=> a-3d+a-d+a+d+a+3d=32

=> 4a=32

=> a=32/4=8

again,

(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

=> 15a^2-7a^2-135d^2+7d^2=0

=> 8a^2-128d^2=0

=>128d^2=8a^2

=>128d^2=8.(8)^2=8*64=512

=> d^2=512/128=4

=> d=+2,-2

so the terms are

(8-3*2),(8-2),(8+2),(8+3*2). [for d=+2]

=2,6,10,14

again ,

the terms are for d= -2

(8-3*-2),{8-(-2)},{8+(-2)},(8+3*-2)

=14,10,6,2