Question

There are n arithmetic means between 1 and 31 such that the 7th mean:(n-1)th mean = 5:9. Find n.​

Answer 1

Answer:

Let the arithmetic mean inserted be A1,A2,A3,....Am.

now , series is 1,A1,A2,A3,....Am,31

hence , a=1,T

n

=31,n=n+2

T

n

=a+(n−1)d

31=1+(n+2−1)d

30=(n+1)d

d=

n+1

30

_______(1)

given that

T

n

−1

T

7

=

9

5

a+(n−1−1+1)d

a+(7−1+1)d

=

9

5

(here , a+(7−1+1)d , +1 after 7-1 is because in the series the 1st A.M. is second in terms of series)

a+(n−1)d

a+7d

=

9

5

9a+63d=5a+5nd−5d

4a+68d=5nd

4a=d(5n−68)

putting value of d frm (1)

by solving this you will get value of n=14