Q. 124.6( 30 Votes )

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 :

Let 32 be divided into parts as (a - 3d), (a - d), (a + d) and (a + 3d).

Now (a - 3d) + (a - d) + (a + d) + (a + 3d) = 32

4a = 32

a = 8

Now, we are given that product of the first and the fourth terms is to the product of the second and the third terms as 7 : 15.

i.e. [(a - 3d) × (a + 3 d)] : [(a - d) × (a + d)] = 7 : 15

=

15[(a - 3d) × (a + 3 d)] = 7[(a - d) × (a + d)]

15[a2 - 9d2] = 7 [a2 - d2]

15a2 - 135d2 = 7a2 - 7d2

8a2 - 128d2 = 0

8a2 = 128d2

Putting the value of a, we get,

512 = 128 d2

d2 = 4

d =  ±2

If d = 2, then the numbers are 2, 6, 10, 14.

If d = - 2, then the numbers are 14, 10, 6, 2.

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