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Question

Between 1 and 31, m arithmetic means are inserted so that the ratio of the 7th and (m−1)th means is 5:9. Then the value of m is

A
12
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B
13
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C
14
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D
15
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Solution

The correct option is C 14
Let the arithmetic mean inserted be a1,a2,a3,...,am.
Now, the series is 1,a1,a2,a3,...,am,31.
Hence a=1,Tn=31,n=m+2.
Now, Tn=a+(n1)d
31=1+(m+21)d
d=30(m+1) (1)
Given T7Tm1=59
a+(71+1)da+(m11+1)d=59
Here, a+(71+1)d, +1 after 71 is because in the series the first A.M is second in the terms of series.
So, a+7da+(m1)d=59
9a+63d=5a+5md5d
4a=d(5m68)
Substitute the values of d and a from equation (1); we get
4(m+1)=30(5m68)
146m=2044
m=14

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