# Class 11 RD Sharma Solutions – Chapter 18 Binomial Theorem- Exercise 18.2 | Set 3

• Last Updated : 16 May, 2021

### Question 27. If the 3rd, 4th, 5th and 6th terms in the expansion of (x + α)n be respectively a, b, c, and d, prove that .

Solution:

We are given, (x + α)n

So, T3 = a = nC2 xn-2 α2

T4 = b = nC3 xn-3 α3

T5 = c = nC4 xn-4 α4

T6 = d = nC5 xn-5 α5

We need to prove that, => => => => => => => => => Hence proved.

### Question 28. If the 6th, 7th, 8th and 9th terms in the expansion of (x + α)n be respectively a, b, c, and d, prove that .

Solution:

We are given, (x + α)n

So, T6 = a = nC5 xn-5 α5

T7 = b = nC6 xn-6 α6

T8 = c = nC7 xn-7 α7

T9 = d = nC8 xn-8 α8

We need to prove that, => => => => => => => => => Hence proved.

### Question 29. If the coefficients of three consecutive terms in the expansion of (1+x)n are respectively 76, 95, and 76, find n.

Solution:

We are given, (1+x)n

Let the three consecutive terms be rth, (r+1)th and (r+2)th.

We know the coefficient of rth term of a binomial expression is given by nCr-1.

Coefficient of rth term = nCr-1 = 76

Coefficient of (r+1)th term = nCr+1-1 = nCr = 95

Coefficient of (r+2)th term = nCr+2-1 = nCr+1 = 76

Now, => => => 5n − 5r = 4r + 4

=> 5n − 9r = 4  . . . . (1)

Also, => => 4n − 4r + 4 = 5r

=> 4n − r = −4  . . . . (2)

Subtracting (2) from (1), we get,

=> n = 4 + 4

=> n = 8

Therefore, the value of n is 8.

### Question 30. If the 6th, 7th, and 8th  in the expansion of (x + a)n be respectively 112, 7, and 1/4, find x, a, and n.

Solution:

We are given, (x + a)n

Also, T6 = nC5 xn-5 a5 = 112

T7 = nC6 xn-6 a6 = 7

T8 = nC7 xn-7 a7 = 1/4

Now, => => => => => . . . . (1)

Also, => => => => => . . . . (2)

From (1) and (2), we get,

=> => => 3n − 18 = 2n − 10

=> n = 8

Putting n = 8 in (2), we get,

=> => => x = 8a

Now, nC5 xn-5 a5 = 112

=> 8C5 x8-5 a5 = 112

=> 8C5 (8a)3 a5 = 112

=> => a8 => a8 => a = 1/2

So, x = 8 (1/2) = 4

Therefore, the value of x, a and n is 4, 1/2 and 8 respectively.

### Question 31. If the 2nd, 3rd, and 4th  in the expansion of (x + a)n be respectively 240, 720, and 1080 respectively, find x, a, and n.

Solution:

We are given, (x + a)n

Also, T2 = nC1 xn-1 a = 240

T3 = nC2 xn-2 a2 = 720

T4 = nC3 xn-3 a3 = 1080

Now, => => => => . . . . (1)

Also, => => => => . . . . (2)

From (1) and (2), we get,

=> => 12n − 24 = 9n − 9

=> 3n = 15

=> n = 5

Putting n = 5 in (2), we get,

=> => => => Now, nC1 xn-1 a = 240

=> 5C1 x5-1 (3x/2) = 240

=> 5C1 x5 (3/2) = 240

=> => => x5 = 32

=> x5 = 25

=> x = 2

So, a = (3/2) (2) = 3

Therefore, the value of x, a and n is 2, 3 and 5 respectively.

### Question 32. Find a, b, and n in the expansion of (a+b)n if the first three terms are 729, 7290, and 30375 respectively.

Solution:

We are given, (a+b)n

Also, T1 = nC0 an = 729

T2 = nC1 an-1 b1 = 7290

T3 = nC2 an-2 b2 = 30375

Now, => => . . . . (1)

Also, => => => => . . . . (2)

From (1) and (2), we get,

=> => 30n − 30 = 25n

=> 5n = 30

=> n = 6

So, nC0 an = 729

=> a6 = 36

=> a = 3

Putting a = 3 in (2), we get,

=> => => b = 5

Therefore, the value of a, b and n is 3, 5 and 6 respectively.

### Question 33. Find a, if the coefficients of x2 and x3 in the expansion of (3+ax)9 are equal.

Solution:

We have, (3+ax)9 = 9C0 39 + 9C1 38 (ax)1 + 9C2 37 (ax)2 + 9C3 36 (ax)3 + . . . .

Coefficient of x2  = 9C2 37 a2

Coefficient of x3  = 9C3 36 a3

According to the question, we have,

=> 9C2 37 a2 = 9C3 36 a3

=> => 81 = 63 a

=> a = 9/7

Therefore, the value of a is 9/7.

### Question 34. Find a, if the coefficients of x and x3 in the expansion of (2+ax)4 are equal.

Solution:

We have, (2+ax)4 = 4C0 24 + 4C1 23 (ax)1 + 4C2 22 (ax)2 + 4C3 2 (ax)3 + . . . .

Coefficient of x  = 4C1 23 a

Coefficient of x3  = 4C3 2 a3

According to the question, we have,

=> 4C1 23 a = 4C3 2 a3

=> 4C3 23 a = 4C3 2 a3

=> 8a = 2a3

=> 2a (a − 4) = 0

=> a = 0 or a = 4

Therefore, the value of a is 0 or 4.

### Question 35. If the term free from x in the expansion of is 405, find the value of k.

Solution:

We have, The general term of this expression will be,

Tr+1  If the term is independent of x , we must have,

=> => 10 − r − 4r = 0

=> 5r = 10

=> r = 2

Therefore, the required term is 3rd term.

So, we have,

=> = 405

=> = 405

=> = 405

=> = 405

=> 45k2 = 405

=> k2 = 9

=> k = 3

Therefore, the value of k is 3.

### Question 36. Find the sixth term in the expansion (y1/2 + x1/3)n, if the binomial coefficient of the third term from the end is 45.

Solution:

We have, (y + x)n

The third term of the expansion from the end is (n + 1 − 3 + 1)th term = (n − 1)th term.

=> Tn-1 = Tn-2+1 = nCn-2 (y1/2)n-(n-2) (x1/3)n-2

The coefficient of this term is given, i.e., 45.

=> nCn-2 = 45

=> n (n − 1)/2 = 45

=> n (n − 1) = 90

=> n2 − n − 90 = 0

=> n2 − 10n + 9n − 90 = 0

=> n(n−10) + 9 (n−10) = 0

=> n = 10 or n = −9 (ignored)

So, the sixth term of the expansion is T6 = T5+1

= 10C10-5 (y1/2)10-(10-5) (x1/3)10-5

= 10C5 (y1/2)5 (x1/3)5

= 252 (y5/2) (x5/3)

### Question 37. If p is a real number and if the middle term in the expansion of (p/2 + 2)8 is 1120, find p.

Solution:

We have, (p/2 + 2)8

Total number of terms is 8 + 1 = 9 (odd number).

The middle term is (9+1)/2 = 5th term.

Therefore, we get T5 = T4+1 = 1120

=> 8C4 (p/2)8-4 (2)4 = 1120

=> 70 (p/2)4 (2)4 = 1120

=> 70p4 = 1120

=> p4 = 16

=> p4 = 24

=> p = 2

Therefore, the value of p is 2.

### Question 38. Find n in the binomial , if the ratio of the 7th term from the beginning to from the end is 1/6.

Solution:

We have, 7th term from the beginning is .

And 7th term from the end is .

According to the question, we have,

=> => => => => (n − 12)/3 = −1

=> n − 12 = −3

=> n = 9

Therefore, the value of n is 9.

### Question 39. If the seventh term from the beginning and end in the binomial expansion of are equal, find n.

Solution:

We have, 7th term from the beginning is .

And 7th term from the end is .

According to the question, we have,

=> => => n − 12 = 12 − n

=> 2n = 24

=> n = 12

Therefore, the value of n is 12.

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