Suppose $p$ is the first of $n (n > 1)$ AM's between two positive numbers $a$ and $b$ , then value of $p$ is

\frac{n a + b}{n + 1}

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\frac{n a - b}{n + 1}

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\frac{n a + a}{n + 1}

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\frac{n a - a}{n + 1}

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Solution

The correct option is A $\frac{n a + b}{n + 1}$
Let $A_{1}, A_{2}, . . . . . A_{n}$ be the n AM's to be inserted between $a$ and $b$
Then, $a, A_{1} . . . . . . . . . . A_{n^{'}}, b$ is an AP. Let the common difference be $d$ .
$b = a + (n + 2 - 1) d$
$\Rightarrow d = \frac{b - a}{n + 1}$
$p = A_{1} = a + \frac{b - a}{n + 1}$
$p = \frac{a n + a + b - a}{n - 1}$
$= \frac{a n + b}{n + 1}$

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