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Combination with Restrictions

If a1,a2,.....

Question

If $a_{1}, a_{2}, . . . a_{n}$ are in HP then the expression $a_{1} a_{2} + a_{2} a_{3} + . . . . a_{n - 1} a_{n}$ equal to

A

a_{1} a_{n} (n - 1)

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B

a_{2} a_{n} (n - 1)

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C

a_{1} a_{n} (n - 2)

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D

2 a_{n} (n - 2)

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Solution

The correct option is A $a_{1} a_{n} (n - 1)$
$a_{1} a_{2} a_{3} \dots a_{n}$ are in $H \cdot P$
So as we Know reverse of $H P$ is $A \cdot P$
$\frac{1}{a_{1}}, \frac{1}{a_{2}}, \frac{1}{a_{3}} - \frac{1}{a_{n}} are in A \cdot P$
$d = \frac{1}{a_{2}} - \frac{1}{a_{1}}$
(common difference of $A P)$
$\Rightarrow a_{1} - a_{2} = a_{1} a_{2} d$
similarly $a_{2} - a_{3} = a_{2} a_{3} d$

$a_{n - 1} - a_{n} = a_{n - 1} a_{n d}$
So now add all of these, we get
$a_{1} - a_{n} = d (a_{1} q_{2} + a_{2} q_{3} + \dots + a_{n}, a_{n}) -$ (i)
Also $\frac{1}{a_{n}} = \frac{1}{a_{1}}, (n - 1) d (n^{th}$ term of $A \cdot p)$
$d = a_{1} - a_{n}$
$a_{1} a_{n} (n - 1)$
Putting value of $d$ in eq 1 we get
$a_{1} - a_{n} = a_{1} - a_{n} (a_{1} a_{2} + a_{2} a_{3} + \dots + a_{n} + q_{n})$
$a_{1} a_{n} (n - 1)$
$a_{1} a_{2} + a_{2} a_{3} + \dots + a_{n} + a_{n} = a_{1} a_{n} (n - 1)$

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