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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 \frac{b^2-ac}{c^2-bd}=\frac{5a}{3c} .

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, \frac{b^2-ac}{c^2-bd}=\frac{5a}{3c}

=> \frac{b^2-ac}{a}=\frac{5(c^2-bd)}{3c}

=> \frac{b^2-ac}{ab}=\frac{5(c^2-bd)}{3bc}

=> \frac{b}{a}-\frac{c}{b}=\frac{5}{3}\left[\frac{c}{b}-\frac{d}{c}\right]      

=> \frac{^nC_3x^{n-3}α^3}{^nC_2x^{n-2}α^2}-\frac{^nC_4x^{n-4}α^4}{^nC_3x^{n-3}α^3}=\frac{5}{3}\left[\frac{^nC_4x^{n-4}α^4}{^nC_3x^{n-3}α^3}-\frac{^nC_5x^{n-5}α^5}{^nC_4x^{n-4}α^4}\right]

=> \left[\frac{^nC_3}{^nC_2}-\frac{^nC_4}{^nC_3}\right]\frac{\alpha}{x}=\frac{5\alpha}{3x}\left[\frac{^nC_4}{^nC_3}-\frac{^nC_5}{^nC_4}\right]

=> \left[\frac{n-2}{3}-\frac{n-3}{4}\right]=\frac{5}{3}\left[\frac{n-3}{4}-\frac{n-4}{5}\right]



=> \frac{4n-8-3n+9}{12}=\frac{5}{3}\left[\frac{5n-15-4n+16}{20}\right]

=> \frac{n+1}{12}=\frac{5}{3}\left[\frac{n+1}{20}\right]

=> \frac{n+1}{12}=\frac{n+1}{12}

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 \frac{b^2-ac}{c^2-bd}=\frac{4a}{3c} .

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, \frac{b^2-ac}{c^2-bd}=\frac{4a}{3c}

=> \frac{b^2-ac}{a}=\frac{4(c^2-bd)}{3c}

=> \frac{b^2-ac}{ab}=\frac{4(c^2-bd)}{3bc}

=> \frac{b}{a}-\frac{c}{b}=\frac{4}{3}\left[\frac{c}{b}-\frac{d}{c}\right]

=> \frac{^nC_6x^{n-6}α^6}{^nC_5x^{n-5}α^5}-\frac{^nC_7x^{n-7}α^7}{^nC_6x^{n-6}α^6}=\frac{4}{3}\left[\frac{^nC_7x^{n-7}α^7}{^nC_6x^{n-6}α^6}-\frac{^nC_8x^{n-8}α^8}{^nC_7x^{n-7}α^7}\right]

=> \left[\frac{^nC_6}{^nC_5}-\frac{^nC_7}{^nC_6}\right]\frac{\alpha}{x}=\frac{4\alpha}{3x}\left[\frac{^nC_7}{^nC_6}-\frac{^nC_8}{^nC_7}\right]

=> \left[\frac{n-5}{6}-\frac{n-6}{7}\right]=\frac{4}{3}\left[\frac{n-6}{7}-\frac{n-7}{8}\right]

=> \frac{7n-35-6n+36}{42}=\frac{4}{3}\left[\frac{8n-48-7n+49}{56}\right]

=> \frac{n+1}{42}=\frac{4}{3}\left[\frac{n+1}{56}\right]

=> \frac{n+1}{42}=\frac{n+1}{42}



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, \frac{^nC_{r+1}}{^nC_r}=\frac{76}{95}

=> \frac{n-(r+1)+1}{r+1}=\frac{76}{95}



=> \frac{n-r}{r+1}=\frac{4}{5}

=> 5n − 5r = 4r + 4

=> 5n − 9r = 4  . . . . (1)

Also, \frac{^nC_{r}}{^nC_{r-1}}=\frac{95}{76}

=> \frac{n-r+1}{r}=\frac{5}{4}

=> 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, \frac{^nC_6x^{n-6}a^6}{^nC_5x^{n-5}a^5}=\frac{7}{112}

=> \frac{^nC_{n-6}x^{n-6}a^6}{^nC_{n-5}x^{n-5}a^5}=\frac{7}{112}

=> \frac{^nC_{n-6}}{^nC_{n-5}}×\frac{a}{x}=\frac{1}{16}

=> \frac{n-6+1}{n-(n-5)+1}×\frac{a}{x}=\frac{1}{16}



=> \frac{n-5}{6}×\frac{a}{x}=\frac{1}{16}

=> \frac{a}{x}=\frac{3}{8(n-5)}      . . . . (1)

Also, \frac{^nC_7x^{n-7}a^7}{^nC_6x^{n-6}a^6}=\frac{\frac{1}{4}}{7}

=> \frac{^nC_{n-7}x^{n-7}a^7}{^nC_{n-6}x^{n-6}a^6}=\frac{1}{28}

=> \frac{^nC_{n-7}}{^nC_{n-6}}×\frac{a}{x}=\frac{1}{28}

=> \frac{n-7+1}{n-(n-6)+1}×\frac{a}{x}=\frac{1}{28}

=> \frac{n-6}{7}×\frac{a}{x}=\frac{1}{28}

=> \frac{a}{x}=\frac{1}{4(n-6)}      . . . . (2)

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

=> \frac{3}{8(n-5)} = \frac{1}{4(n-6)}



=> \frac{3}{2(n-5)} = \frac{1}{n-6}

=> 3n − 18 = 2n − 10

=> n = 8

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

=> \frac{a}{x}=\frac{1}{4(8-6)}

=> \frac{a}{x}=\frac{1}{8}

=> x = 8a

Now, nC5 xn-5 a5 = 112

=> 8C5 x8-5 a5 = 112

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

=> \frac{8×7×6}{3×2}×512×a^3×a^5 = 112

=> a8\frac{1}{256}

=> a8(\frac{1}{2})^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 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, \frac{^nC_3x^{n-3}a^3}{^nC_2x^{n-2}a^2}=\frac{1080}{720}

=> \frac{^nC_3}{^nC_2}×\frac{a}{x}=\frac{3}{2}

=> \frac{n-3+1}{2+1}×\frac{a}{x}=\frac{3}{2}

=> \frac{n-2}{3}×\frac{a}{x}=\frac{3}{2}

=> \frac{a}{x}=\frac{9}{2(n-2)}     . . . . (1)

Also, \frac{^nC_2x^{n-2}a^2}{^nC_1x^{n-1}a}=\frac{720}{240}

=> \frac{^nC_2}{^nC_1}×\frac{a}{x}=3

=> \frac{n-2+1}{2}×\frac{a}{x}=3

=> \frac{n-1}{2}×\frac{a}{x}=3



=> \frac{a}{x}=\frac{6}{n-1}      . . . . (2)

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

=> \frac{9}{2(n-2)}=\frac{6}{n-1}

=> 12n − 24 = 9n − 9

=> 3n = 15

=> n = 5

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

=> \frac{a}{x}=\frac{6}{5-1}

=> \frac{a}{x}=\frac{6}{4}

=> \frac{a}{x}=\frac{3}{2}

=> a = \frac{3}{2}x

Now, nC1 xn-1 a = 240

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

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

=> \frac{5×4!}{4!}×x^5×\frac{3}{2} = 240

=> x^5=\frac{240×2}{5×3}

=> 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, \frac{^nC_1a^{n-1}b^1}{^nC_0a^n} = \frac{7290}{729}

=> ^nC_{n-1}×\frac{b}{a} = 10

=> \frac{b}{a} = \frac{10}{n}       . . . . (1)

Also, \frac{^nC_2a^{n-2}b^2}{^nC_1a^{n-1}b} = \frac{30375}{7290}



=> \frac{^nC_{n-2}}{^nC_{n-1}}×\frac{b}{a} = \frac{25}{6}

=> \frac{n-2+1}{n-(n-1)+1}×\frac{b}{a} = \frac{25}{6}

=> \frac{n-1}{2}×\frac{b}{a} = \frac{25}{6}

=> \frac{b}{a} = \frac{25}{3(n-1)}     . . . . (2)

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

=> \frac{10}{n} = \frac{25}{3(n-1)}

=> 30n − 30 = 25n 

=> 5n = 30

=> n = 6

So, nC0 an = 729

=> a6 = 36

=> a = 3

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

=> \frac{b}{3} = \frac{25}{3(6-1)}

=> \frac{b}{3} = \frac{5}{3}

=> 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

=> \frac{9×8}{2×1}×3=\frac{9×8×7}{3×2×1}×a

=> 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 (\sqrt{x}-\frac{k}{x^2})^{10}    is 405, find the value of k.

Solution:

We have, (\sqrt{x}-\frac{k}{x^2})^{10}

The general term of this expression will be,

Tr+1^{10}C_r(\sqrt{x})^{10-r}(-\frac{k}{x^2})^{r}

^{10}C_r.(x)^{\frac{10-r}{2}-2r}(-k)^r

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

=> \frac{10-r}{2}-2r=0

=> 10 − r − 4r = 0

=> 5r = 10

=> r = 2

Therefore, the required term is 3rd term.

So, we have, 

=> ^{10}C_2(\sqrt{x})^{10-2}(-\frac{k}{x^2})^{2}    = 405

=> ^{10}C_2(\sqrt{x})^{8}(\frac{k^2}{x^4})    = 405

=> ^{10}C_2(x^{4})(\frac{k^2}{x^4})    = 405 

=> \frac{10×9}{2×1}×k^2    = 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 (\sqrt[2]2+\frac{1}{\sqrt[3]3})^n , if the ratio of the 7th term from the beginning to from the end is 1/6.

Solution:

We have, (\sqrt[2]2+\frac{1}{\sqrt[3]3})^n

7th term from the beginning is ^nC_6(\sqrt[3]2)^{n-6}(\frac{1}{\sqrt[3]3})^6   .

And 7th term from the end is ^nC_{n-6}(\sqrt[3]2)^{6}(\frac{1}{\sqrt[3]3})^{n-6}   .

According to the question, we have,

=> \frac{^nC_6(\sqrt[3]2)^{n-6}(\frac{1}{\sqrt[3]3})^6}{^nC_{n-6}(\sqrt[3]2)^{6}(\frac{1}{\sqrt[3]3})^{n-6}}=\frac{1}{6}

=> (\sqrt[3]2)^{n-12}(\frac{1}{\sqrt[3]3})^{12-n}=\frac{1}{6}

=> (\sqrt[3]2)^{n-12}(\sqrt[3]3)^{n-12}=\frac{1}{6}

=> (6)^{\frac{n-12}{3}}=6^{-1}

=> (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 (\sqrt[2]2+\frac{1}{\sqrt[3]2})^n  are equal, find n.

Solution:



We have, (\sqrt[2]2+\frac{1}{\sqrt[3]2})^n

7th term from the beginning is ^nC_6(\sqrt[3]2)^{n-6}(\frac{1}{\sqrt[3]2})^6   .

And 7th term from the end is ^nC_{n-6}(\sqrt[3]2)^{6}(\frac{1}{\sqrt[3]2})^{n-6}   .

According to the question, we have,

=> ^nC_6(\sqrt[3]2)^{n-6}(\frac{1}{\sqrt[3]2})^6 = \hspace{0.1cm}^nC_{n-6}(\sqrt[3]2)^{6}(\frac{1}{\sqrt[3]2})^{n-6}

=> (\sqrt[3]2)^{n-12}=(\sqrt[3]2)^{12-n}

=> n − 12 = 12 − n

=> 2n = 24

=> n = 12

Therefore, the value of n is 12.

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