Question

There are n A.M.s between 3 and 17. The ratio of the last mean to the first mean is 3: 1. Find the value of n.

Answer 1

Let the series be 3, A₁, A₂, A₃, …….., A_n, 17

Given, a_n/a₁ = 3/1

We know total terms in AP are n + 2

So, 17 is the (n + 2)th term

By using the formula,

A_n = a + (n – 1)d

A_n = 17, a = 3

So, 17 = 3 + (n + 2 – 1)d

17 = 3 + (n + 1)d

17 – 3 = (n + 1)d

14 = (n + 1)d

d = 14/(n + 1)

Now,

A_n = 3 + 14/(n + 1) = (17n + 3) / (n + 1)

A₁ = 3 + d = (3n + 17)/(n + 1)

Since,

a_n/a₁ = 3/1

(17n + 3)/ (3n + 17) = 3/1

17n + 3 = 3(3n + 17)

17n + 3 = 9n + 51

17n – 9n = 51 – 3

8n = 48

n = 48/8

= 6

∴ There are 6 terms in the AP