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If n arithmetic mean are inserted between 1 and 31 such that the ratio of the first mean is 3:29 find the value of n?
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If n arithmetic mean are inserted between 1 and 31 such that the ratio...
a = 1  a(n+2) = 31

given, a2 / a (n +1) = 3 / 29  ----i   { as A1 will be 2nd term of AP and An will be a(n+1) term}

a2 - a = a(n+2) - a(n+1)

a + d - a  = 31 - a - nd

d = 31 - 1 - nd

d+nd = 30

d(1+n) = 30

d = 30/n+1

from i

a + d / a + nd = 3/29

29 + 29d = 3 + 3nd

26 = 3nd - 29d

26 = d(3n - 29)

26 = (3n - 29) 30 / n+1

26n + 26 = 90n - 870

90n - 26n = 896

64n = 896

Therefore n =14

Hence,  14  A.M.'s are inserted between  1 and 31.
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If n arithmetic mean are inserted between 1 and 31 such that the ratio of the first mean is 3:29 find the value of n?
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