Question

Find four consecutive terms in A.P whose sum is 72 and the ratio of product of the extreme terms to the product of means is 9:10

Answer 1

the four terms are

a, a+d, a+2d, a+3d

their sum = 72
4a + 6d = 72
2a + 3d = 36 **

product of extremes: a(a+3d)
product of means: (a+d)(a+2d)

a(a+3d) / (a+d)(a+2d) = 9/10
10a^2 + 30ad = 9a^2 + 27ad + 18d^2
a^2 + 3ad - 18d^2 = 0
(a + 6d)(a - 3d) = 0
a = -6d or a = 3d ***

if a = 3d, in **
2(3d) + 3d = 36
9d = 36
d = 4 , then a = 12
<b>and the terms are 12, 16, 20, 24</b>

if a = -6d in **
2(-6d) + 3d = 36
-9d = 36
d = -4 , a = 24
terms are : 24, 20, 16, 12
notice the terms are just in reverse order

Answer 2

four consecutive terms are a-3d , a-2d, a+2d , a+3d

Answer 3

sb high school jalna

Answer 4

excellent

Answer 5

Thanks a lot!

Answer 6

333³33333333333