If a 1 - a 2 - a 3 - a n are in AP - then 1 a 1 a 2 - 1 a 2 a 3 - 1 a 3 a 4 - - - 1 a n - 1 a n equals-

The correct option is B n 1 a _ 1 a _ n a_1,a_2,...........a_n are in APLet common difference= d then>a_2 a_1=d a_3 a_2=d>e ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Sum of n Terms

If a 1 · a...

Question

If $a_{1} \cdot a_{2} \cdot a_{3} \dots \dots a_{n}$ are in AP , then $\frac{1}{a_{1} a_{2}} + \frac{1}{a_{2} a_{3}} + \frac{1}{a_{3} a_{4}} + \dots \dots + \frac{1}{a_{n - 1} a_{n}}$ equals-

\frac{a_{1} a_{2}}{n - 1}

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

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

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

\frac{n - 1}{a_{1} - a_{n}}

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

\frac{n - 1}{a_{1} a_{n}}

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

Open in App

Solution

Suggest Corrections

Similar questions

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}}

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

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

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

If $a_{1}, a_{2}, a_{3}, . . . ., a_{n}$ be an AP of non-zero terms. Prove that

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

$\frac{n - 1}{a_{1} a_{n}}$

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