First slide

Means - A.M/G.M/H.M

Question

Between two numbers whose sum is 2 $\frac{1}{6}$ , an even number of arithmetic means are inserted. If the sum of these means exceeds their number by unity, then the number of means are

Moderate

A

12
B

10
C

8
D

None of these

Solution

Let 2n arithmetic means be A1, A2, A3, …, A2n between a and b.
$Then A_{1} + A_{2} + A_{3} + \dots + A_{2 n} = \frac{a + b}{2} \times 2 n$
$= \frac{\frac{13}{6}}{2} \times 2 n = \frac{13 n}{6}$
Given, $A_{1} + A_{2} + A_{3} + \dots + A_{2 n} = 2 n + 1$
$\begin{array}{ll} ∴ 2 n + 1 = \frac{13 n}{6}; or 12 n + 6 = 13 n \\ ∴ n = 6 \end{array}$
$∴ The number of means = 2 n = 2 \times 6 = 12 .$

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