Given that n A.M.'s are inserted between two sets of numbers a, 2b and 2a, b, where a, b ∈ R. If the mth means in the two cases are same then ratio a: b is equal to:-
(1) n : (n – m + 1) (2) (n – m + 1) : m (3) (n – m + 1) : n (4) m : (n – m + 1)
Hi,
Let their common difference be d. As there are n AMs between a & b, total number of terms in the sequence is = n + 2
==> nth term b = a + (n + 1)d; ==> d = (b - a)/(n + 1)
So, 2nd term, that is First mean A₍1₎ = a + [(b - a)/(n + 1)]
3rd term that is Second mean A₍2₎ = a + [2*(b - a)/(n + 1)]
In this way rth mean = a + [r*(b - a)/(n + 1)]
Solution:
i) In the first sequence, first term is = a and the nth term is 2b and there are n AMs between them. As such from the above concept,
rth mean is: a + [r*(2b - a)/(n + 1)]
Hence the mth mean is: a + [m*(2b - a)/(n + 1)]
ii) Similarly for the second sequence, mth mean = 2a + [m*(b - 2a)/(n + 1)]
iii) Since the mth means are equal, equating both the above,
a + [m*(2b - a)/(n + 1)] = 2a + [m*(b - 2a)/(n + 1)]
==> [m*(2b - a)/(n + 1)] - [m*(b - 2a)/(n + 1)] = 2a - a
==> {m/(n + 1)}*(2b - a - b + 2a) = a
==> (a + b)/a = (n + 1)/m
Subtracting 1 on both sides of the above,
b/a = (n + 1 - m)/m
So, a/b = m/(n - m + 1)
Regards
Let their common difference be d. As there are n AMs between a & b, total number of terms in the sequence is = n + 2
==> nth term b = a + (n + 1)d; ==> d = (b - a)/(n + 1)
So, 2nd term, that is First mean A₍1₎ = a + [(b - a)/(n + 1)]
3rd term that is Second mean A₍2₎ = a + [2*(b - a)/(n + 1)]
In this way rth mean = a + [r*(b - a)/(n + 1)]
Solution:
i) In the first sequence, first term is = a and the nth term is 2b and there are n AMs between them. As such from the above concept,
rth mean is: a + [r*(2b - a)/(n + 1)]
Hence the mth mean is: a + [m*(2b - a)/(n + 1)]
ii) Similarly for the second sequence, mth mean = 2a + [m*(b - 2a)/(n + 1)]
iii) Since the mth means are equal, equating both the above,
a + [m*(2b - a)/(n + 1)] = 2a + [m*(b - 2a)/(n + 1)]
==> [m*(2b - a)/(n + 1)] - [m*(b - 2a)/(n + 1)] = 2a - a
==> {m/(n + 1)}*(2b - a - b + 2a) = a
==> (a + b)/a = (n + 1)/m
Subtracting 1 on both sides of the above,
b/a = (n + 1 - m)/m
So, a/b = m/(n - m + 1)
Regards