If the coefficients of $r^{t h}, (r + 1)^{t h} a n d (r + 2)^{t h}$ terms in the binomial expansion of $(1 + y)^{m}$ are in A.P., then $m$ and $r$ satisfy the equation

m^{2} - m (4 r - 1) + 4 r^{2} + 2 = 0

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

m^{2} + m (4 r + 1) + 4 r^{2} + 2 = 0

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

m^{2} - m (4 r + 1) + 4 r^{2} - 2 = 0

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

m^{2} - m (4 r - 1) - 4 r^{2} + 2 = 0

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

Open in App

Solution

The correct option is C $m^{2} - m (4 r + 1) + 4 r^{2} - 2 = 0$ .
Given binomial expansion is $(1 + y)^{m}$

$T_{r} = T_{r - 1 + 1} =^{m} C_{r - 1} x^{r - 1}$

$T_{r + 1} =^{m} C_{r} x^{r}$

$T_{r + 2} = T_{r + 1 + 1} =^{m} C_{r + 1} x^{r + 1}$
Since, $^{m} C_{r - 1},^{m} C_{r},^{m} C_{r + 1}$ are in A.P.

$\Rightarrow 2^{m} C_{r} =^{m} C_{r - 1} +^{m} C_{r + 1}$
$\Rightarrow 2 = \frac{^{m} C_{r - 1}}{^{m} C_{r}} + \frac{^{m} C_{r + 1}}{_{^{m} C_{r}}}$
$\Rightarrow 2 = \frac{r}{m - r + 1} + \frac{m - r}{r + 1}$ $(∵ \frac{^{n} C_{r}}{^{n} C_{r - 1}} = \frac{n - r + 1}{r})$
$\Rightarrow m^{2} - m (4 r + 1) + 4 r^{2} - 2 = 0$

Suggest Corrections

Similar questions

If the coefficients of $r^{t h}, (r + 1)^{t h}$ and $(r + 2)^{t h}$ terms in the binomial expansion of $(1 + y)^{m}$ are in $A . P .$ , then $m$ and $r$ will satisfy the equation

Q. If the coefficient of

r^{t h}, (r + 1)^{t h}

and

(r + 2)^{t h}

terms in the binomial expansion of

{(1 + y)}^{n}

are in A.P, then

n^{2} - n (4 r + 1) + {4 r}^{2} - 2

is equal to

Q. If the coefficient of

r^{t h}, (r + 1)^{t h}

and

(r + 2)^{t h}

terms in the expansion of

(1 + x)^{n}

are in A.P., then show that

n^{2} - (4 r + 1) n + 4 r^{2} - 2 = 0

Q. If the coefficients of

r^{t h}, {(r + 1)}^{t h}, {(r + 2)}^{t h}

terms in

{(1 + x)}^{n}

are in A.P. then show that

n^{2} - (4 r + 1) n + 4 r^{2} - 2 = 0

If the coefficients of $x^{r - 1}, x^{r}$ and $x^{r + 1}$ in the binomial expansion of $(1 + x)^{n}$ are in AP, prove that
$n^{2} - n (4 r + 1) + 4 r^{2} - 2 = 0$

Join BYJU'S Learning Program

If the coefficients of rth,(r+1)th and (r+2)th terms in the binomial expansion of (1+y)m are in A.P., then m and r satisfy the equation

The correct option is C m2−m(4r+1)+4r2−2=0.Given binomial expansion is (1+y)mTr=Tr−1+1=mCr−1xr−1Tr+1=mCrxrTr+2=Tr+1+1=mCr+1xr+1Since, mCr−1,mCr,mCr+1 are in A.P.⇒2mCr=mCr−1+mCr+1⇒2=mCr−1mCr+mCr+1mCr⇒2=rm−r+1+m−rr+1 (∵nCrnCr−1=n−r+1r)⇒m2−m(4r+1)+4r2−2=0

If the coefficients of $r^{t h}, (r + 1)^{t h} a n d (r + 2)^{t h}$ terms in the binomial expansion of $(1 + y)^{m}$ are in A.P., then $m$ and $r$ satisfy the equation