Suppose m arithmetic means are inserted between 1 and 31. If the ratio of the second mean to the mth mean is 1 : 4, then mis equal to

Let m arithmetic means between 1 and 31 be

$A_{1}, A_{2}, \dots, A_{m},$ then $1, A_{1}, A_{2} \dots A_{m}$ 31 is an A.P. Let $d$ be

the common difference of this A.P. Now, $31 = (m + 2)$ th term of the A.P. Now,

$= 1 + (m + 2 - 1) d$

$\Rightarrow$ $d = \frac{30}{m + 1} .$

Also, $A_{2} = 1 + 2 d, A_{m} = 1 + m d = 31 - d$

Now, $\frac{A_{2}}{A_{m}} = \frac{1}{4} \Rightarrow 4 A_{2} = A_{m}$

$\Rightarrow 4 (1 + 2 d) = 31 - d \Rightarrow d = 3$

Thus $\frac{30}{m + 1} = 3 \Rightarrow m = 9 .$

Suppose $m$ arithmetic means are inserted between 1 and 31. If the ratio of the second mean to the $m$ th mean is 1 : 4, then $m$
is equal to