Answer :

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

(a^{2} – 9d^{2})(a^{2} – d^{2}) = 945

Putting a = 6

We get,

(36– 9d^{2})(36 – d^{2}) = 945

1296 – 36d^{2} – 324d^{2} + 9d^{4} = 945

351 – 360d^{2} + 9d^{4} = 0

9d^{4} – 360d^{2} + 351 = 0

d^{4} – 40d^{2} + 39 = 0

d^{4} – 39d^{2} – d^{2 +} 39 = 0

d^{2}(d^{2} – 39) – 1 (d^{2} – 39) = 0

d^{2} – 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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