Question

Find three number in AP such that their sum is 60and last one is three
time the first

Answer 1

Since we know that sequence of three terms in A.P. is a - d, a and a + d.

→ a + d = 3(a - d)

→ a + d = 3a - 3d

→ 2a = 4d

→ a = 2d

→ a + d + a + a - d = 60

→ 3a = 60

→ a = 20

Hence, a = 2d or a/2 = d

→ d = 20/2 = 10

Hence we have an A.P.,

→ 10, 20, 30

Answer 2

GIVEN:

Some of 3 terms in an AP = 60

The last term of an AP an = 3

Let the terms be (a - d), a and (a + d)

a - d + a + a + d = 60

=> 3a = 60

=> a = 60/3

=> a = 20

The last term is 3 times the first term

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

a + d = 3a - 3d

d + 3d = 3a - a

=> 4d = 2a

=> 4d = 2(20)

=> 4d = 40

=> d = 40/4

=> d = 10

The terms are:

First term = a - d = 20 - 10 = 10

Second term = a = 20

The third term = a + d = 20 + 10 = 30

Therefore, the terms are 10, 20 and 30.