Question

The sum of n A.M.'s between two numbers is 20. If the last mean is double of the first and one number is three times the other, then find the numbers.

Answer 1

Answer:

The numbers are 2 & 6.

Step-by-step explanation:

Here,

Let the numbers are a & b and d be a common difference.

From the last case,

b = 3a.................(i)

Again, from the second case,

mn = 2m1

or, b - d = 2(a + d)

or, 3a - d = 2a + 2d

or, a = 3d........................(ii)

Now,  

mn = a + nd

or, b - d = 3d + nd  

or, 9d - d = 3d + nd         [ Since, b= 3a & a = 3d.]

or, 5d = nd

or, n = 5.

Again,  

From the first case,

Sum of AMs = 20

or, n/2 x (m1 + mn) = 20           [ Sum = n/2 x [ first term + last term.]

or, 5(4d + 8d) = 40

or, d = 2/3.

So, a = 3d = 3 x 2/3 = 2 and  

b = 3a = 3 x 2 = 6.

Thus the numbers are 2 & 6.