Question

Given 4 numbers are in 3 are in G. P. & last 3 are in A.P. Sum of extremes=32 & that of mid-terms is = 24 Find the numbers​

Answer 1

Step-by-step explanation:

Sum of 4 terms of an AP = 32.

Let the terms be a-d, a, a+d, and a+2d.

The sum is: 4a+2d=32, or

2a+d=16 …(1)

It says (a-d)(a+2d): a(a+d)::7:15, or

15 (a-d)(a+2d) = 7a(a+d), or

15(a^2+ad-2d^2) = 7a^2+7ad, or

15a^2 + 15ad - 30d^2 = 7a^2+7ad, or

8a^2+8ad-30d^2 = 0, or

4a^2+4ad-15d^2 = 0, or

4a^2+10ad-6ad-15d^2 = 0, or

2a(2a+5d)-3d(2a+5d)=0, or

(2a-3d)(2a+5d)=0

So a = 3d/2 or -5d/2 …(3)

Again 2a+d=16 …(1) or

a =(16-d)/2 …(2)

From (2) and (3)

3d/2 =(16-d)/2, or

3d=16-d, or

4d=16, or d = 4. So a = 6 …(4)

Also -5d/2 =(16-d)/2, or

-5d+d = 16 or -4d=16 or d= -4. So a =10. …(5)

So the 4 terms of the AP are 2, 6, 10 and 14 OR 14, 10, 6 and 2.

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