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

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, $A_2=1+2d, A_m=1+md=31-d$

Now, $\frac{A_2}{A_m}=\frac{1}{4}\Rightarrow 4A_2=A_m$

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

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