Question

Divide 32 into four parts which are the four terms of an AP such that the product of the first and the fourth terms is to the product of the second and the third terms as 7 : 15.

Answer 1

Let the four parts in AP be (a - 3d),(a - d),(a + d),(a + 3d).

Given, Sum of four parts = 32.

⇒ a - 3d + a - d + a + d + a + 3d = 32

⇒ 4a = 32

⇒ a = 8.

Now,

Given : Product of extremes to product of means is 7:15.

⇒ (a - 3d)(a + 3d)/(a - d)(a + d) = 7/15

⇒ (8 - 3d)(8 + 3d)/(8 - d)(8 + d) = 7/15

⇒ 15(8 - 3d)(8 + 3d) = 7(8 - d)(8 + d)

⇒ 15(64 - 9d^2) = 7(64 - d^2)

⇒ 960 - 135d^2 = 448 - 7d^2

⇒ 128d^2 = 512

⇒ d^2 = 4

⇒ d = 2, -2.

When a = 8 and d = 2:

(i) a - 3d = 2

(ii) a - d = 6

(iii) a + d = 10

(iv) a + 3d = 14.

When a = 8 and d = -2:

(i) a - 3d = 14

(ii) a - d = 10

(iii) a + d = 6

(iv) a + 3d = 2.

Therefore, the four terms of the AP is 2,6,10,14.

Hope this helps!

Answer 2

Hey there !!


▶ Question :-

→ Divide 32 into four parts which are the four terms of an AP such that the product of the first and the fourth terms is to the product of the second and the third terms as 7 : 15.


▶ Solution :-

→ Given :-

➡ The product of the first and the fourth terms is to the product of the second and the third terms as 7 : 15.


For Further solution :-


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▶ Identity used :-

→ ( a + b ) ( a - b ) = a² - b².


✔✔ Hence, it is solved ✅✅.

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