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

Answer:

Step-by-step explanation:

Let the four parts be (a−3d),(a−d),(a+d) and (a+3d) which are in A.P.

Given that the sum of all parts of a number is 32.

⇒(a−3d)+(a−d)+(a+d)+(a+3d)=32⇒4a=32⇒a=324⇒a=8

Now, we have to find the value of d by using the ratio of the product of extremes to the product of means is 7:15.

Product of extreme =(a−3d)(a+3d)

Product of means =(a−d)(a+d)

Now, the ratio of the product of extremes to the product of means is 7:15.

⇒(a−3d)(a+3d)(a−d)(a+d)=715

Using the identity (x+y)(x−y)=x2−y2

⇒a2−9d2a2−d2=715

Apply cross multiplication,

⇒15(a2−9d2)=7(a2−d2)⇒15a2−135d2=7a2−7d2⇒135d2−7d2=15a2−7a2⇒128d2=8a2

Now put the value of a=8

⇒d2=8×(8)2128⇒d2=512128⇒d2=4

Taking square root,

⇒d=±2

Case 1: a=8 and d=2

(a−3d)=8−3×2=2(a−d)=8−2=6(a+d)=8+2=10(a+3d)=8+3×2=14

Four parts are 2, 6, 10 and 14.

Case 2: a=8 and d=-2

(a−3d)=8−3×(−2)=14(a−d)=8−(−2)=10(a+d)=8+(−2)=6(a+3d)=8+3×(−2)=2

Four parts are 14, 10, 6 and 2.

So, the four parts are 2, 6 , 10 and 14.

Answer 2

Answer:

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