# Find four numbers in AP such that their sum is 24 and product is 945.

Let the four terms of AP be a – 3d, a – d, a + d, a + 3d

Given, sum = 24

(a – 3 d ) + (a – d) + (a + d ) + (a + 3 d) = 24

4a = 24

a = 6

Product = 945

(a2 – 9d2)(a2 – d2) = 945

Putting a = 6

We get,

(36– 9d2)(36 – d2) = 945

1296 – 36d2 – 324d2 + 9d4 = 945

351 – 360d2 + 9d4 = 0

9d4 – 360d2 + 351 = 0

d4 – 40d2 + 39 = 0

d4 – 39d2 – d2 + 39 = 0

d2(d2 – 39) – 1 (d2 – 39) = 0

d2 – 1 = 0

d = ±1

d = ±√39

So, terms of AP are 3, 5, 7, 9

As with d = √39 product will not be 945

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