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

Question 27. If the 3^rd, 4^th, 5^th and 6^th 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 x^n-2 α²

T₄ = b = ⁿC₃ x^n-3 α³

T₅ = c = ⁿC₄ x^n-4 α⁴

T₆ = d = ⁿC₅ x^n-5 α⁵

We need to prove that, 

=> 

=> 

=>    

=> 

=> 

=> 

=> 

=> 

=> 

Hence proved.

Question 28. If the 6^th, 7^th, 8^th and 9^th 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 x^n-5 α⁵

T₇ = b = ⁿC₆ x^n-6 α⁶

T₈ = c = ⁿC₇ x^n-7 α⁷

T₉ = d = ⁿC₈ x^n-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 r^th, (r+1)^th and (r+2)^th.

We know the coefficient of r^th term of a binomial expression is given by ⁿC_r-1.

Coefficient of r^th term = ⁿC_r-1 = 76

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

Coefficient of (r+2)^th term = ⁿC_r+2-1 = ⁿC_r+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 6^th, 7^th, and 8^th  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 x^n-5 a⁵ = 112

T₇ = ⁿC₆ x^n-6 a⁶ = 7

T₈ = ⁿC₇ x^n-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 x^n-5 a⁵ = 112

=> ⁸C₅ x^8-5 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 2^nd, 3^rd, and 4^th  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 x^n-1 a = 240

T₃ = ⁿC₂ x^n-2 a² = 720

T₄ = ⁿC₃ x^n-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 x^n-1 a = 240

=> ⁵C₁ x^5-1 (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 a^n-1 b¹ = 7290

T₃ = ⁿC₂ a^n-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,

T_r+1 = 

= 

If the term is independent of x , we must have,

=> 

=> 10 − r − 4r = 0

=> 5r = 10

=> r = 2

Therefore, the required term is 3^rd 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 (y^1/2 + x^1/3)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. 

=> T_n-1 = T_n-2+1 = ⁿC_n-2 (y^1/2)^n-(n-2) (x^1/3)^n-2 

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

=> ⁿC_n-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

= ¹⁰C_10-5 (y^1/2)^10-(10-5) (x^1/3)^10-5 

= ¹⁰C₅ (y^1/2)⁵ (x^1/3)⁵

= 252 (y^5/2) (x^5/3)

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, 

7^th term from the beginning is .

And 7^th 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, 

7^th term from the beginning is .

And 7^th 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.