Question

Between 1 and 31, m numbers have been inserted in such a way that the resulting sequence is an A.P. and the ratio of 7 th and ( m – 1) th numbers is 5:9. Find the value of m .

Answer 1

The numbers have been inserted between 1 and 31. The ratio of 7 th and ( m−1 ) th term is 5:9.

Let A 1 , A 2 , A 3 ,……, A m be the m numbers, such that 1, A 1 , A 2 , A 3 ,……, A m ,31 is an A.P.

The formula for the common difference while introducing arithmetic mean between two terms is given by,

d= b−a n+1 (1)

Here,

a=1 b=31 n=m

Substitute the values in equation (1).

d= 31−1 m+1 = 30 m+1 (2)

The 7 th term of the sequence is given by,

A 7 =a+7d

Similarly, the ( m−1 ) th term of the sequence is given by,

A m−1 =a+( m−1 )d

As per the given condition,

A 7 A m−1 = 5 9

Substitute the values of A 7 and A m−1 in the above expression,

a+7d a+( m−1 )d = 5 9 (3)

Substitute the value of d from equation (2) to equation (3).

1+7( 30 m+1 ) 1+( m−1 )( 30 m+1 ) = 5 9 1+7( 30 m+1 ) 1+( m−1 )( 30 m+1 ) = 5 9 m+1+7( 30 ) m+1+30( m−1 ) = 5 9 m+1+210 m+1+30m−30 = 5 9

Further simplify the above expression.

m+211 31m−29 = 5 9 ( m+210 )×9=( 31m−29 )×5 9m+1899=155m−145 155m−9m=1899+145

Further simplify the above expression.

146m=2044 m= 2044 146 =14

Thus, the value of m is 14.