Q2 of 8 Page 180

Find four numbers in AP such that their sum is 24 and product is 945.

Let the four terms of AP be a – 3d, a – d, a + d, a + 3d


Given, sum = 24


(a – 3 d ) + (a – d) + (a + d ) + (a + 3 d) = 24


⇒ 4a = 24


a = 6


Product = 945


(a2 – 9d2)(a2 – d2) = 945


Putting a = 6


We get,


(36– 9d2)(36 – d2) = 945


1296 – 36d2 – 324d2 + 9d4 = 945


351 – 360d2 + 9d4 = 0


9d4 – 360d2 + 351 = 0


d4 – 40d2 + 39 = 0


d4 – 39d2 – d2 + 39 = 0


d2(d2 – 39) – 1 (d2 – 39) = 0


d2 – 1 = 0


d = ±1


d = ±√39


So, terms of AP are 3, 5, 7, 9


As with d = √39 product will not be 945


More from this chapter

All 8 →
1

If in an A.P the sum of m terms is equal to n and the sum of n terms is equal to m then prove that the sum of (m – n) terms is – (m + n).

 3

The sum of the first 20 terms of a G.P is 244 times the sum of its first 10 terms. Find the common ratio.

 4

In a set of four numbers, the first three are in G.P. and the last three are in A.P. with a common difference of 6. If the first number is the same as the fourth, find the four numbers.

 5

Three numbers are in G.P whose sum is 70. If the extreme is each multiplied by 4 and the mean by 5 they will be in A.P. Find the numbers.