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Standard VIII

Mathematics

Factorisation by Common Factors

Let A1,A2,....

Question

Let $A_{1}, A_{2}, . . . A_{n} a n d H_{1}, H_{2}, . . . H_{n}$ are $n$ $A . M s$ & $H . M s$ between the two quantities $a$ & $b$ $(a, b > 0)$ respectively. On the basis is of above information answer the following questionsThe value of $\frac{A_{n} - A_{1}}{b - a}$ is equal to

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

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

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

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

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Solution

The correct option is C $\frac{n - 1}{n + 1}$
$a, A_{1}, A_{2}, . . ., A_{n}, b$ are in AP
$b = a + (n + 1) d$
$\Rightarrow d = \frac{b - a}{n + 1}$
$A_{1} = a + d, A_{n} = a + n d$
$A_{n} - A_{1} = (n - 1) d$
$= (n - 1) \frac{b - a}{n + 1}$
$∴ \frac{A_{n} - A_{1}}{b - a} = \frac{n - 1}{n + 1}$

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Let A1,A2,...An and H1,H2,...Hn are n A.Ms & H.Ms between the two quantities a & b (a,b>0) respectively. On the basis is of above information answer the following questionsThe value of An−A1b−a is equal to

The correct option is C n−1n+1a,A1,A2,...,An,b are in APb=a+(n+1)d⇒d=b−an+1A1=a+d,An=a+nd An−A1=(n−1)d =(n−1)b−an+1∴An−A1b−a=n−1n+1

Let $A_{1}, A_{2}, . . . A_{n} a n d H_{1}, H_{2}, . . . H_{n}$ are $n$ $A . M s$ & $H . M s$ between the two quantities $a$ & $b$ $(a, b > 0)$ respectively. On the basis is of above information answer the following questionsThe value of $\frac{A_{n} - A_{1}}{b - a}$ is equal to

The correct option is C $\frac{n - 1}{n + 1}$
$a, A_{1}, A_{2}, . . ., A_{n}, b$ are in AP
$b = a + (n + 1) d$
$\Rightarrow d = \frac{b - a}{n + 1}$
$A_{1} = a + d, A_{n} = a + n d$
$A_{n} - A_{1} = (n - 1) d$
$= (n - 1) \frac{b - a}{n + 1}$
$∴ \frac{A_{n} - A_{1}}{b - a} = \frac{n - 1}{n + 1}$