Maths · Insertion of arithmetic, geometric means between two given numbers
If arithmetic means are inserted between 1 and 31 such that ratio of first and m
If \( n \) arithmetic means are inserted between 1 and 31 such that ratio of first and \( n^{t h} \) mean is \( 3: 29, \) then what is the value of \( n ? \)
- A. 10
- B. 14
- C. 18
- D. 23
Step-by-step solution
Let n arithmetic means be inserted between 1 and 31. The sequence becomes an AP with first term 1, last term 31, and (n+2) terms. Common difference d = (31-1)/(n+1)=30/(n+1). First mean A1 = 1+d, nth mean An = 1+nd. Given ratio A1:An = 3:29, so (1+d)/(1+nd)=3/29. Cross-multiplying: 29+29d=3+3nd → 26=3nd-29d = d(3n-29). Substitute d: 30/(n+1)=26/(3n-29) → 30(3n-29)=26(n+1) → 90n-870=26n+26 → 64n=896 → n=14.