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Suppose arithmetic means are inserted between 1 and 31. If the ratio of the second mean to the th mean is 1 : 4, then
is equal to
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a
7
b
9
c
11
d
15
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Correct option is B
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.Similar Questions
Between 1 and 31 are inserted M arithmetic means so that the ratio of the 7th and (M - 1) th means is 5:9, then the value of m is
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