If $a_{1}, a_{2}, a_{3}, . . . a_{n}$ are in HP then $a_{1} a_{2} + a_{2} a_{3} + a_{3} a_{4} + . . . + a_{n - 1} a_{n} =$

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

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

(n + 4) 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 D $(n - 1) a_{1} a_{n}$
If $a_{1}, a_{2}, a_{3} . . . .$ are in H.P then $\frac{1}{a_{1}}, \frac{1}{a_{2}}, \frac{1}{a_{3}} . . .$ are in A.P.

$\frac{1}{a_{n}} = \frac{1}{a_{1}} + (n - 1) d$

$d = \frac{(a_{1} - a_{n})}{[a_{1} \times a_{n} (n - 1)]}$

Now we know $\frac{1}{a_{2}} - \frac{1}{a_{1}} = d$

$a_{1} \times a_{2} = \frac{a_{1} - a_{2}}{d}$

Similarly $a_{2} \times a_{3} = \frac{a_{2} - a_{3}}{d}$
So $a_{1} a_{2} + a_{2} a_{3} + a_{3} a_{4} . . . = \frac{(a_{1} - a_{2})}{d} + \frac{(a_{2} - a_{3})}{d} + \frac{(a_{3} - a_{4})}{d} . . . + \frac{(a_{n - 1} - a_{n})}{d}$
All terms will be cancelled except the first and the $n^{t h}$ term so expression becomes $\frac{(a_{1} - a_{n})}{d}$
On substituting value of d we get the expression equal to $a_{1} \times a_{n} \times (n - 1)$

Suggest Corrections

Similar questions

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 $a_{1}, a_{2}, a_{3}, \dots, a_{n}$ be an AP of nonzero terms then prove that

$\frac{1}{a_{1} a_{2}} + \frac{1}{a_{2} a_{3}} + \frac{1}{a_{3} a_{4}} + \dots + \frac{1}{a_{n - 1} a_{n}} = \frac{(n - 1)}{a_{1} a_{n}}$

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

\frac{1}{a_{1} a_{2}} + \frac{1}{a_{2} a_{3}} + \frac{1}{a_{3} a_{4}} + . . . . . + \frac{1}{a_{n - 1} a_{n}}

a_{1}, a_{2}, a_{3}, a_{4}, a_{5}

a_{1} a_{2} + a_{2} a_{3} + a_{3} a_{4} + a_{4} a_{5}

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

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

Join BYJU'S Learning Program