If $a_{1}, a_{2}, . . . a_{n}$ are in H.P. then the expression $a_{1} a_{2} + a_{2} a_{3} + . . . + a_{n - 1} a_{n} i s e q u a l t o$

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

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

n a_{1} a_{n}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $(n - 1) a_{1} a_{n}$
$a_{1}, a_{2}, a_{3} \cdot \dots a_{n}$ are in $H P$
$∴ \frac{1}{a_{1}}, \frac{1}{a_{2}}, \frac{1}{a_{3}} \cdot \cdot \frac{1}{a_{n}}$ are in $A \cdot P$
$\frac{1}{a_{2}} - \frac{1}{a_{1}} = d \Rightarrow \frac{a_{1} - a_{2}}{a_{1} a_{2}} = d$
$a_{1} a_{2} = \frac{a_{1} - a_{2}}{d}$
$a_{1} a_{2} = \frac{a_{1} - a_{2}}{d}$
Similarly $, a_{2} a_{3} = \frac{a_{2} - a_{3}}{d} \cdot \dots \cdot a_{n - 1} a_{n} = \frac{a_{n - 1} - a_{n}}{d}$
$a_{1} a_{2} + a_{2} a_{3} + \dots \cdot a_{n - 1} a_{n} = \frac{a_{1} - a_{2}}{d} + \frac{a_{2} - a_{3}}{d} + \dots + \frac{a_{n - 1} - a_{n}}{d}$
$= \frac{1}{d} (a_{1} - α_{2} + a_{2} - \dots + a_{n - 1} - a_{n})$
$= \frac{a_{1} - a_{n}}{d}$
To find "d
$\frac{1}{a_{n}} = \frac{1}{a_{1}} + (n - 1) d$
$\Rightarrow \frac{1}{a_{n}} = \frac{1}{a_{1}} + n d - d$
$\Rightarrow \frac{1}{a_{n}} - \frac{1}{a_{1}} = d (n - 1)$
$\Rightarrow \frac{a_{1} - a_{n}}{(a_{1} a_{n}) (n - 1)} = d$
$∴ a_{1} a_{2} + a_{2} a_{3} + \dots \cdot a_{n - 1} a_{n} = \frac{a_{1} - a_{n}}{d}$
$= \frac{a_{1} - a_{n}}{\frac{a_{1} - a_{n}}{(a_{1} a_{n}) (n - 1)}}$
$= \frac{(n - 1) (a_{1} - α_{n}) (a_{1} a_{n})}{(a_{1} - g_{n})}$
$= (n - 1) (a_{1} a_{n})$
$∴$ Answer: option $(A)$

Suggest Corrections

Similar questions

Given that a_i $>$ 0 and i belongs to a set of natural numbers. If $a_{1}, a_{2}, a_{3} . . . . . a_{2 n}$ are in AP, then find the value of $\frac{a_{1} + a_{2 n}}{\sqrt{a_{1}} + \sqrt{a_{2}}} + \frac{a_{2} + a_{2 n} - 1}{{\sqrt{a}}_{2} + \sqrt{a_{3}}} . . . . . . . . . . . . + \frac{a_{n} + a_{n + 1}}{\sqrt{a_{n}} + \sqrt{a_{n + 1}}}$