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 7th and (m – 1)th numbers is 5:9. Find the value of m.

Answer 1

Since m numbers are inserted between 1 and 31 then the total number of terms in the A.P. = n = (m+2).
The first term = a = 1
The last term = $t_{m+2}$
= a+(m+2-1)d= 31
or, 1+(m+1)d=31
or, (m+1)d=30
or, d=30/(m+1) ---------------------(1)
Now, according to the question,
 [a+7d]/[a+(m-1)d]=5/9
or, (1+7d)/{1+(m-1)d}=5/9
or, 9+63d=5+5(m-1)d
or, 63d-5(m-1)d=5-9
or,(63-5m+5)d=-4
or, (68-5m){30/(m+1)}=-4
or, 30(68-5m)=-4(m+1)
or, 2040-150m=-4m-4
or, -150m+4m=-2040-4
or, -146m=-2044
or, m=(-2044)/(-146)
or, m=14 Ans.

Answer 2

Answer:

your the answer is following :