Question

the sum of 4 consecutive terms of an ap is32 and the ratio of the product of the first and last term is to the product of middle 2 term is 7:15 find the number
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Answer 1

take first 4 terms a-3d ,a-d ,a+3d a+d

then adding up we get 4a=32

a=8

Answer 2

FOR 4 NOS. IN AP THEY MUST BE ARRANGED IN

:(a+3d),(a+d),(a-d),(a-3d)

ATQ

SUM OF ALL 4 CONSECUTIVE NUMBERS

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

=4a=32

=a=8

case 2

(a+3d) (a-3d)/(a+d) (a-d) =7/15

CROSS MULTIPLICATION

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

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

8a^2=128d^2

128 *4 /128 =d^2

4=d^2

2=d

SO, THE AP IS AS FOLLOWS:14,10,6,2