Question

find 4 consecutive term of a.p whose sum is 66 and product of extremists to product of means is 14:15

Answer 1

HELLO DEAR,



assume the terms to be (a-3d), (a-d), (a+d)
and (a+3d)


a-3d +a-d+a+d+a+3d= 66

4a = 66


a= 66/4= 33/2


since product of extremes to the product of
means is 14/15



(a-3d)(a+3d)/(a-d)(a+d) = 14/15



a2 - 9d2/ a2- d2 = 14/15


15 ( a2- 9d​2) = 14 ( a2- d2)


15a2- 135d2= 14a2- 14d2


15a2-14a2= 135d2-14d2


a2= 121d2


substituting value of a,


(33/2)2= 121d2


1089/4 = 121d2


d2= 1089/4*121


d= 1.5


so the ap is,


a-3d

= >16.5-4.5= 12



a-d= 16.5- 1.5= 15



a+d = 16.5 + 1.5 = 18



a+3d = 16.5 + 4.5= 21


the ap is = 12,15,18,21

I HOPE ITS HELP YOU DEAR

THANKS