Question

guys plz solve this
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:)

Answer 1

take the four parts of the A.P as a-3d,a-d,a+d and a+3d then solve it

Answer 2

Let four numbers in A.P. be a – 3d, a – d, a + d and a + 3d.
(a – 3d) + (a – d) + (a + d) + (a + 3d) = 32
⇒ 4a = 32
⇒ a = 8
Product of extremes = (a – 3d) (a + d) + (8 – 3d) (8 + 3d) = 64 – 9d2
Product of means = (a – d) (a + d) = (8 – d) (8 + d) = 64 – d2

Product of extremes: product of means = 7:15 (Given)


When a = 8 and d = 2
a – 3d = 8 – 3 × 2 = 8 – 6 = 2
a – d = 8 – 2 = 6
a + d = 8 + 2 = 10
a + 3d = 8 + 3 × 2 = 8 + 6 = 14

When a = 8, d = –2
a – 3d = 8 – 3 × (–2) = 8 + 6 = 14
a – d = 8 – (–2) = 8 + 2 = 10
a + d = 8 – 2 = 6
a + 3d = 8 + 3 × (–2) = 8 – 6 = 2

∴ The four numbers are 2, 6, 10 and 14.