Sir pls solve this question.
Asked by rsudipto
13th December 2018,
10:09 AM
Answered by Expert
Answer:
n number of arithmetic means are inserted between a and 2b. Hence in A.P., 2b is (n+2)th term
2b = a + (n+1) d ................(1)
where d is the common difference.
Also we are given that, mth mean, i.e., (m+1)th term in A.P., is given by n = a+md .............(2)
We eliminate d in eqn.(1), using eqn.(2), 2b = a +(n+1)(n-a)/m or 2mb = (m-n-1)a + n(n+1) ............(3)
Similarly, n number of arithmetic means are inserted between 2a and b. Hence in A.P., b is (n+2)th term
b = 2a + (n+1) d ................(4)
where d is the common difference.
Also we are given that, mth mean, i.e., (m+1)th term in A.P., is given by n = 2a+md .............(5)
We eliminate d in eqn.(5), using eqn.(5), b = 2a +(n+1)(n-2a)/m or mb = (m-n-1)2a + n(n+1) ............(6)
By solving the simultaneous linear equations (3) and (6), we get, 
Hence ratio of a to b = m : ( n - m + 1)
Answered by Expert
13th December 2018,
4:43 PM
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