# 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** **.**

^{rd}, 4

^{th}, 5

^{th}and 6

^{th}terms in the expansion of (x + α)

^{n}be respectively a, b, c

**Solution:**

We are given, (x + α)

^{n}So, T

_{3}= a =^{n}C_{2}x^{n-2}α^{2}T

_{4}= b =^{n}C_{3}x^{n-3}α^{3}T

_{5}= c =^{n}C_{4}x^{n-4}α^{4}T

_{6}= d =^{n}C_{5}x^{n-5}α^{5}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 ****.**

^{th}, 7

^{th}, 8

^{th}and 9

^{th}terms in the expansion of (x + α)

^{n}be respectively a, b, c

**Solution:**

We are given, (x + α)

^{n}So, T

_{6}= a =^{n}C_{5}x^{n-5}α^{5}T

_{7}= b =^{n}C_{6}x^{n-6}α^{6}T

_{8}= c =^{n}C_{7}x^{n-7}α^{7}T

_{9}= d =^{ n}C_{8}x^{n-8}α^{8}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.**

^{n}are respectively 76, 95

**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^{n}C_{r-1}.Coefficient of r

^{th}term =^{n}C_{r-1}= 76Coefficient of (r+1)

^{th}term =^{n}C_{r+1-1}=^{n}C_{r}= 95Coefficient of (r+2)

^{th}term =^{n}C_{r+2-1}=^{n}C_{r+1}= 76Now,

=>

=>

=> 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.**

^{th}, 7

^{th}

^{th}in the expansion of (x + a)

^{n}be respectively 112, 7

**Solution:**

We are given, (x + a)

^{n}Also, T

_{6 }=^{n}C_{5}x^{n-5}a^{5}= 112T

_{7}=^{n}C_{6}x^{n-6}a^{6}= 7T

_{8}=^{n}C_{7}x^{n-7}a^{7}= 1/4Now,

=>

=>

=>

=>

=> . . . . (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,

^{n}C_{5}x^{n-5}a^{5}= 112=>

^{8}C_{5}x^{8-5}a^{5}= 112=>

^{8}C_{5}(8a)^{3}a^{5}= 112=>

=> a

^{8}==> a

^{8}==> 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.**

^{nd}, 3

^{rd}

^{ }and 4

^{th}in the expansion of (x + a)

^{n}be respectively 240, 720

**Solution:**

We are given, (x + a)

^{n}Also, T

_{2}=^{n}C_{1 }x^{n-1}a = 240T

_{3}=^{n}C_{2}x^{n-2}a^{2}= 720T

_{4}=^{n}C_{3}x^{n-3}a^{3}= 1080Now,

=>

=>

=>

=> . . . . (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,

^{ n}C_{1}x^{n-1}a = 240=>

^{5}C_{1}x^{5-1}(3x/2) = 240=>

^{5}C_{1}x^{5}(3/2) = 240=>

=>

=> x

^{5}= 32=> x

^{5}= 2^{5}=> 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.**

^{n}if the first three terms are 729, 7290

**Solution:**

We are given, (a+b)

^{n}Also, T

_{1}=^{n}C_{0}a^{n}= 729T

_{2}=^{n}C_{1}a^{n-1}b^{1}= 7290T

_{3}=^{n}C_{2}a^{n-2}b^{2}= 30375Now,

=>

=> . . . . (1)

Also,

=>

=>

=>

=> . . . . (2)

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

=>

=> 30n − 30 = 25n

=> 5n = 30

=> n = 6

So,

^{n}C_{0}a^{n}= 729=> a

^{6}= 3^{6}=> 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 x**^{2} and x^{3} in the expansion of (3+ax)^{9} are equal.

^{2}and x

^{3}in the expansion of (3+ax)

^{9}are equal.

**Solution:**

We have, (3+ax)

^{9 }=^{9}C_{0}3^{9}+^{9}C_{1}3^{8}(ax)^{1}+^{9}C_{2}3^{7}(ax)^{2 }+^{9}C_{3}3^{6}(ax)^{3}+ . . . .Coefficient of x

^{2}=^{9}C_{2}3^{7}a^{2}Coefficient of x

^{3}=^{9}C_{3}3^{6}a^{3}According to the question, we have,

=>

^{9}C_{2}3^{7}a^{2}=^{9}C_{3}3^{6 }a^{3}=>

=> 81 = 63 a

=> a = 9/7

Therefore, the value of a is 9/7.

**Question 34. Find a, if the coefficients of x and x**^{3} in the expansion of (2+ax)^{4} are equal.

^{3}in the expansion of (2+ax)

^{4}are equal.

**Solution:**

We have, (2+ax)

^{4}=^{4}C_{0}2^{4 }+^{4}C_{1}2^{3}(ax)^{1}+^{4}C_{2}2^{2}(ax)^{2}+^{4}C_{3 }2 (ax)^{3}+ . . . .Coefficient of x =

^{4}C_{1}2^{3}aCoefficient of x

^{3}=^{4}C_{3}2 a^{3}According to the question, we have,

=>

^{4}C_{1}2^{3}a =^{4}C_{3}2 a^{3}=>

^{4}C_{3}2^{3}a =^{4}C_{3}2 a^{3}=> 8a = 2a

^{3}=> 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

=> 45k

^{2}= 405=> k

^{2 }= 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.

^{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}=^{n}C_{n-2}(y^{1/2})^{n-(n-2)}(x^{1/3})^{n-2 }The coefficient of this term is given, i.e., 45.

=>

^{n}C_{n-2}= 45=> n (n − 1)/2 = 45

=> n (n − 1) = 90

=> n

^{2}− n − 90 = 0=> n

^{2}− 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 T

_{6 }= T_{5+1}=

^{10}C_{10-5 }(y^{1/2})^{10-(10-5)}(x^{1/3})^{10-5}=

^{10}C_{5}(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.

^{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 T

_{5}= T_{4+1 }=_{ }1120=>

^{8}C_{4}(p/2)^{8-4}(2)^{4}= 1120=> 70 (p/2)

^{4}(2)^{4}= 1120=> 70p

^{4 }= 1120=> p

^{4}= 16=> p

^{4 }= 2^{4}=> 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.

Attention reader! Don’t stop learning now. Participate in the **Scholorship Test for First-Step-to-DSA Course for Class 9 to 12 students**.