If n arithmetic mean are inserted between 1 and 31 such that the ratio of the first mean and nth mean is 3:29,find the value of n
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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
.:n =14
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
.:n =14
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