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 beA1, A2,…, Am, then 1, A1, A2…Am 31 is an A.P. Let d bethe common difference of this A.P. Now, 31=(m+2)th term of the A.P. Now,=1+(m+2 –1)d ⇒ d=30m+1.Also, A2=1+2d, Am=1+md=31−dNow, A2Am=14⇒4A2=Am⇒ 4(1+2d)=31−d⇒d=3Thus 30m+1=3⇒m=9.