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

(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 - 1}

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

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

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

(n + 1) a_{1} a_{n + 1}

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

Open in App

Solution

The correct option is A $(n - 1) a_{1} a_{n}$
$a_{1,} a_{2,} a_{3} . . . a_{n} \to H . P$

$∴ \frac{1}{a_{1}}, \frac{1}{a_{2}} . . . \frac{1}{a_{n}} \to A . P$

Let first term be $a$ and common difference be $d$

$∴ \frac{1}{a_{1}} = a \frac{1}{a_{2}} = a + d, \frac{1}{a_{n}} = a + (n - 1) d$

Now

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

$= \frac{1}{(a) (a + d)} + \frac{1}{(a + d) (a + 2 d)} + \frac{1}{(a + 2 d) (a + 3 d)} + . . . \frac{1}{(a + (n - 2) d) (a + (n - 1) d)}$

$= \frac{1}{d} [\frac{d}{a (a + d)} + \frac{d}{(a + d) (a + 2 d)} + \frac{d}{(a + 2 d) (a + 3 d)} + . . . + \frac{d}{(a + (n - 2) d) (a + (n - 1) d)}]$

$= \frac{1}{d} [\frac{1}{a} - \frac{1}{a + d} + \frac{1}{a + d} - \frac{1}{a + 2 d} + . . . . + \frac{1}{a + (n - 2) d} - \frac{1}{a - (n - 1) d}]$

$= \frac{1}{d} [\frac{1}{a} - \frac{1}{a + (n - 1) d}] = \frac{1}{d} [\frac{a + (n - 1) d - a}{(a) (a + (n - 1) d)}]$

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

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

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

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

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

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

Join BYJU'S Learning Program