CameraIcon

SearchIcon

MyQuestionIcon

1

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

General Term of Binomial Expansion

Let nϵ N.If...

Question

Let $n ϵ N$ .If $(1 + x)^{n} = a_{0} x + a_{1} x + a_{2} x^{2} + . . . + a_{n} x^{n},$ and $a_{n - 3}, a_{n - 2}, a_{n - 1},$ are in AP then

A

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

are in

A P

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

B

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

are in

H P

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

C

n = 7

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

D

n = 14

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

Open in App

Solution

The correct options are
A $a_{1}, a_{2}, a_{3}$ are in $A P$
C $n = 7$
$a_{n - 3}, a_{n - 2}, a_{n - 1}$ are in A.P indicates that
$a_{3}, a_{2}, a_{1}$ are in A.P.
Since they are binomial coefficients.
For the above terms to be in A.P, they must follow the relationship
$(n - 2 r)^{2} = n + 2$
Here r will be 2.
Therefore
$n^{2} - 8 r + 16 = n + 2$
$n^{2} - 9 r + 14 = 0$
$(n - 7) (n - 2) = 0$
$n = 7$ since $n = 2$ is not possible here.

Suggest Corrections

thumbs-up

0

Similar questions

n ϵ N

{(1 + x)}^{n} = a_{0} + a_{1} x + a_{2} x^{2} + \cdot \cdot + a_{n} x^{n}

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

{(1 + x)}^{n} = 1 + a_{1} x + a_{1} x^{2} + . . . . + a_{n} x^{n}

a_{1}

a_{2}

a_{3}

are in A.P., then the value of n is

a_{1}, a_{2}, a_{3} \dots, a_{n}

are in AP, where

a_{i} > 0

for all i, then the value of

\frac{1}{\sqrt{a_{1}} + \sqrt{a_{2}}} + \frac{1}{\sqrt{a_{2}} + \sqrt{a_{3}}} + \dots + \frac{1}{\sqrt{a_{n - 2}} + \sqrt{a_{n}}}

a_{r} > 0, r ϵ N

a_{1}

a_{2}

a_{3}

a_{2 n}

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

a_{r} > 0; \forall r, n \in N

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

are in A.P, then

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

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

General Term

MATHEMATICS

Watch in App

explore_more_icon

Explore more

General Term of Binomial Expansion

Standard XII Mathematics

Join BYJU'S Learning Program

CrossIcon