If n arithmetic means are inserted between a and 100 such that the ratio of the first mean to the last mean is 1:7 and a + n = 33, then the value of n is

(c) Let A_1, A_2, ... , A_n be n arithmetic means \ between a and 100. \ The sequence is a, A_1, A_2, ... , A_n, 100. \ The common difference is d = (100-a / n+1). \ Given the ratio of the first mean to \ the last mean is 1:7, so (A_1 / A_n) = (1 / 7). \ => (a+d / a+nd) = (1 / 7) \ => fraca + (100-a / n+1)a + (n(100-a) / n+1) = (1 / 7) \ => (an+a+100-a / an+a+100n-an) = (1 / 7) \ => (an+100 / a+100n) = (1 / 7) \ => 7an + 700 = a + 100n \ It is given that a+n=33 => a = 33-n. \ => 7(33-n)n + 700 = (33-n) + 100n \ => 231n - 7n^2 + 700 = 33 + 99n \ => 7n^2 - 132n - 667 = 0 \ Solving for n: \ => n = 23, -(29 / 7) \ (Ignore negative value as n represents a count) \ Therefore n = 23 \

If n arithmetic means are inserted between a and 100 s... | Sequence & Series (Mathematics)

\text{(c) Let } A_1, A_2, \ldots, A_n \text{ be } n \text{ arithmetic means } \\ \text{between } a \text{ and } 100. \\\\ \text{The sequence is } a, A_1, A_2, \ldots, A_n, 100. \\\\ \text{The common difference is } d = \frac{100-a}{n+1}. \\\\ \text{Given the ratio of the first mean to } \\ \text{the last mean is} 1:7, \text{ so } \frac{A_1}{A_n} = \frac{1}{7}. \\\\ \begin{aligned} &\Rightarrow \frac{a+d}{a+nd} = \frac{1}{7} \\\\ &\Rightarrow \frac{a + \frac{100-a}{n+1}}{a + \frac{n(100-a)}{n+1}} = \frac{1}{7} \\\\ &\Rightarrow \frac{an+a+100-a}{an+a+100n-an} = \frac{1}{7} \\\\ &\Rightarrow \frac{an+100}{a+100n} = \frac{1}{7} \\\\ &\Rightarrow 7an + 700 = a + 100n \end{aligned} \\\\ \text{It is given that } a+n=33 \Rightarrow a = 33-n. \\\\ \begin{aligned} &\Rightarrow 7(33-n)n + 700 = (33-n) + 100n \\\\ &\Rightarrow 231n - 7n^2 + 700 = 33 + 99n \\\\ &\Rightarrow 7n^2 - 132n - 667 = 0 \\\\ &\text{Solving for } n: \\ &\Rightarrow n = 23, \quad -\frac{29}{7} \\ &\text{(Ignore negative value as } n \text{ represents a count)} \\ &\therefore n = 23 \end{aligned}

Explanation

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