Question

five positive numbers are in AP the sum of three middle terms is 24 and the product of first and fifth term is 48. Find the terms of AP

Answer 1

let the 5 +ve integers be
a-3d,a-d,a,a+d,a+3d
sum of middle 3 No's
a-d+a+a+d=24
3a=24
=>a=8
product of first n last term
(a-3d)×(a+3d)=48
a^2-(3d)^2=48
(8^2) - (3^2)(d^2) = 48
64-9d^2=48
-9d^2=48-64
-9d^2 = -16
d^2=16/9
d=4/3
so the required terms of the AP are,,
a-3d=8-3(4/3)=4
a-d=8-4/3=20/3
a=8
a+d=8+4/3=28/3
a+3d=8+3(4/3)=12
therefore, the AP is......
4, 20/3, 8, 28/3, 12
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Hope it helps you....

Answer 2

≡QUESTION≡

Five positive integer are in A.P the sum of middle 3 term is 24 and product of first and fifth term 48. Find the term of A.P.

                                                     

║⊕ANSWER⊕║

Series be a−2d, a−d, a, a+d, a+2d

The sum of first 3 terms

$a - d+a+a+d=24\\3a=24\\a = 8$

$(a+2d)(a-2d) = 48$

$a^{2} - 4d^{2} = 48$

Substitute a = 8

$64 - 4d^{2} = 48\\d^{2} = 16\\d = 2 \\ d = -2$

∴ The AP is 4,6,8,10,12