Question

If the sum of three numbers in a.p is 24 and their product is 440, find the numbers.

Answer 1

Answer:

(a-d) +a+(a+d)=24

3a=24

a=24/3

a=8

(a-d)*a*(a+d)=440

(8-d)*(8+d)=440/8=55

8^2-d^2=55

64-d^2=55

d^2=64-55

d^2=9

d=3 / -3

a-d=8-3= 5 or a-d = 8-(-3)=11

a=8

a+d = 8+3=11 or a+d = 8-3 = 5

So the numbers are 11,8,5 or 5,8,11

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Answer 2

Let’s assume the terms in the A.P to be (a – d), a, (a + d) with common difference as d.

Given conditions,

Sn = 24

(a – d) + a + (a + d) = 3a = 24

a = 24/3 = 8

And,

Product of terms = 440

(a – d) x a x (a + d) = 440

a(a2 – d2) = 440

8(64 – d2) = 440

(64 – d2) = 55

d2 = 9

d = ± 3

Hence,

When a = 8 and d = 3, we have

A.P. = 5, 8, 11

And, when a = 8 and d = -3 we have

A.P. = 11, 8, 5