Question

❤️❤️❤️❤️HELLO FRIENDS ❤️❤️❤️

please solve this question.

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Answer 1

Let the four consecutive numbers in AP be a,a + d, a + 2d, a + 3d.

Given that Sum of four consecutive numbers is 32.

⇒ a + a + d + a + 2d + a + 3d = 32

⇒ 4a + 6d = 32

⇒ 2a + 3d = 16    ------- (1)

Now,

Product of first and last terms to the product of two middle terms is 7:15.

⇒ (a)(a + 3d)/(a + d)(a + 2d) = 7/15

⇒ 15(a^2 + 3ad) = 7(a^2 + 2ad + ad + 2d^2)

⇒ 15(a^2 + 3ad) = 7(a^2 + 3ad + 2d^2)

⇒ 15a^2 + 45ad = 7a^2 + 21ad + 14d^2

⇒ 8a^2 + 24ad - 14d^2 = 0

⇒ 4a^2 + 12ad - 7d^2 = 0

⇒ 4a^2 + 14ad - 2ad - 7d^2 = 0

⇒ 2a(2a + 7d) - d(2a + 7d) = 0

⇒ (2a - d)(2a + 7d) = 0

⇒ 2a = d (or) 2a = -7d.

Substitute 2a = d in (1), we get

⇒ 2a + 3d = 16

⇒ 4d = 16

⇒ d = 4.

Substitute d = -7d in (1), we get

⇒ 2a + 3d = 16

⇒ -7d + 3d = 16

⇒ -4d = 16

⇒ d = -4.

When d = -4:

⇒ 2a + 3d = 16

⇒ 2a + 3(-4) = 16

⇒ 2a - 12 = 16

⇒ 2a = 28

⇒ a = 14.

When d = 4:

⇒ 2a + 3d = 16

⇒ 2a + 12 = 16

⇒ 2a = 4

⇒ a = 2.

So, the four consecutive numbers are 14,10,6,2 (or) 2,6,10,14.

Hope this helps!