If $a_{1}, a_{2}, . . . ., a_{n}$ are in H.P., then the expression $a_{1} a_{2} + a_{2} a_{3} + . . . . + a_{n - 1} a_{n}$

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

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

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

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

Open in App

Solution

The correct option is D $(n - 1) a_{1} a_{n}$
Given that, $a_{1}, a_{2}, a_{3}, \dots, b_{n} ϵ H . P .$
$\Rightarrow \frac{1}{a_{1}}, \frac{1}{a^{2}}, . . . \frac{1}{a_{n}} ϵ A . P .$
$∴ d = \frac{a_{1} - a_{2}}{a_{1} a_{2}} = \frac{a_{2} - a_{3}}{a_{2} a_{3}} = . . . = \frac{a_{n - 1} - a_{n}}{a_{n - 1} a_{n}} = \frac{a_{1} - a_{n}}{(n - 1) a_{1} a_{n}}$
$∴ d = \frac{a_{1} - a_{n}}{a_{1} a_{2} + a_{2} a_{3} + . . . + a_{n - 1} a_{n}}$
$\Rightarrow \frac{a_{1} - a_{n}}{(n - 1) a_{1} a_{n}} = \frac{a_{1} a_{n} (\frac{1}{a_{n}} - \frac{1}{a_{1}})}{a_{1} a_{2} + a_{2} a_{3} + . . . + a_{n - 1} a_{n}}$
Therefore, $a_{1} a_{2} + a_{2} a_{3} + . . . + a_{n - 1} a_{n} = (n - 1) a_{1} a_{n}$
Ans: D

Suggest Corrections

Similar questions

a_{1}, a_{2}, a_{3} . . a_{n}

a_{1} a_{2} + a_{2} a_{3} + . . + a_{n - 1} a_{n} =

a_{1}, a_{2}, . . . a_{n}

a_{1} a_{2} + a_{2} a_{3} + . . . . a_{n - 1} a_{n}

a_{1}, a_{2}, a_{3}, . . . a_{n}

a_{1} a_{2} + a_{2} a_{3} + a_{3} a_{4} + . . . . + a_{n - 1} a_{n} =

If $a_{1}, a_{2}, a_{3}$ ,................., $a_{n}$ are in H.P., then $a_{1} a_{2} + a_{2} a_{3} +$ ...........+ $a_{n - 1} a_{n}$ will be equal to

a_{1}, a_{2}, . . ., a_{n}

d

a_{1} a_{2} + a_{2} a_{3} + . . . . . + a_{n - 1} a_{n}

Join BYJU'S Learning Program