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(C) Let the arithmetic means be $A_1, A_2, \dots, A_n$. The common difference is $d = \frac{100 - a}{n + 1}$.The first mean is $A_1 = a + d$ and the l

Vedclass · Accepted Answer

$23$

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(C) Let the arithmetic means be $A_1, A_2, \dots, A_n$. The common difference is $d = \frac{100 - a}{n + 1}$.The first mean is $A_1 = a + d$ and the last mean is $A_n = 100 - d$.Given $\frac{A_1}{A_n} = \frac{1}{7}$,we have $\frac{a + d}{100 - d} = \frac{1}{7}$.Cross-multiplying gives $7(a + d) = 100 - d$,which simplifies to $7a + 8d = 100$.Substituting $d = \frac{100 - a}{n + 1}$,we get $7a + 8\left(\frac{100 - a}{n + 1}\right) = 100$.Given $a + n = 33$,we substitute $a = 33 - n$ into the equation:$7(33 - n) + 8\left(\frac{100 - (33 - n)}{n + 1}\right) = 100$$231 - 7n + 8\left(\frac{67 + n}{n + 1}\right) = 100$$131 - 7n + \frac{536 + 8n}{n + 1} = 0$$(131 - 7n)(n + 1) + 536 + 8n = 0$$131n + 131 - 7n^2 - 7n + 536 + 8n = 0$$-7n^2 + 132n + 667 = 0 \Rightarrow 7n^2 - 132n - 667 = 0$.Solving the quadratic equation: $(n - 23)(7n + 29) = 0$.Since $n$ must be a positive integer,$n = 23$.

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

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