If n arithmetic means are inserted between 20 and 80 such that the ratio of first mean to the last mean is 1:3, then the value of n is

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Solution

The correct option is C 11
Let $A_{1}, A_{2}, A_{3}, A_{4},, . . . ., A_{n}$ be n arithmetic means between 20 and 80 and d be the common difference of the AP thus formed $20, A_{1}, A_{2}, A_{3}, A_{4},, . . . ., A_{n}, 80 If m AM' s are inserted between a and b, then Common Difference d = \frac{b - a}{m + 1} d = \frac{80 - 20}{n + 1} = \frac{60}{n + 1} A_{1} = 20 + d = 20 + \frac{60}{n + 1} = \frac{20 n + 80}{n + 1} A_{n} = 20 + nd = 20 + n \frac{60}{n + 1} = \frac{80 n + 20}{n + 1} \frac{A_{1}}{A_{n}} = \frac{1}{3} \Rightarrow \frac{\frac{20 n + 80}{n + 1}}{\frac{80 n + 20}{n + 1}} = \frac{1}{3} \Rightarrow \frac{20 (n + 4)}{20 (4 n + 1)} = \frac{1}{3} \Rightarrow 3 n + 12 = 4 n + 1 \Rightarrow n = 11$

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