Maths

Find $n$ in the binomial expansion $,$ if the ratio of the 7th term from the beginning to the 7th term from the end is $\frac{1}{6} .$

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Contributor-Level 10

This is a Long Answer Type Questions as classified in NCERT Exemplar
Sol:

$\begin{array}{l} T h e g i v e n expression i s {(+ \frac{1}{})}^{n} \\ = {(2^{1 / 3} + \frac{1}{3^{1 / 3}})}^{n} \\ G e n e r a l T e r m T_{r + 1} = {}_{n}{C_{r}} x^{n - r} y^{r} \\ T_{7} = T_{6 + 1} = {}_{n}{C_{6}} {(2^{\frac{1}{3}})}^{n - 6} {(\frac{1}{3^{\frac{1}{3}}})}^{6} \\ = {}_{n}{C_{6}} {(2)}^{\frac{n - 6}{3}} . (\frac{1}{3^{2}}) = {}_{n}{C_{6}} {(2)}^{\frac{n - 6}{3}} . {(3)}^{- 2} \\ 7 t h t e r m f r o m t h e e n d = {(n - 7 + 2)}^{t h} t e r m f r o m t h e b e g i n n i n g \\ = {(n - 5)}^{t h} t e r m f r o m t h e b e g i n n i n g \\ S o, T_{n - 6 + 1} = {}_{n}{C_{n - 6}} {(2^{\frac{1}{3}})}^{n - n + 6} {(\frac{1}{3^{\frac{1}{3}}})}^{n - 6} \\ = {}_{n}{C_{n - 6}} {(2)}^{2} . (\frac{1}{3^{\frac{n - 6}{3}}}) = {}_{n}{C_{n - 6}} {(2)}^{2} . {(3)}^{\frac{6 - n}{3}} \\ A c c o r d i n g t o t h e q u e s t i o n, w e g e t \\ \frac{{}_{n}{C_{6}} {(2)}^{\frac{n - 6}{3}} . {(3)}^{- 2}}{{}_{n}{C_{n - 6}} {(2)}^{2} . {(3)}^{\frac{6 - n}{3}}} = \frac{1}{6} \\ \Rightarrow \frac{{}_{n}{C_{n - 6}} {(2)}^{\frac{n - 6}{3}} . {(3)}^{- 2}}{{}_{n}{C_{n - 6}} {(2)}^{2} . {(3)}^{\frac{6 - n}{3}}} = \frac{1}{6} \Rightarrow {(2)}^{\frac{n - 6}{3} - 2} . {(3)}^{- 2 - \frac{6 - n}{3}} = \frac{1}{6} \\ \Rightarrow {(2)}^{\frac{n - 6 - 6}{3}} . {(3)}^{\frac{- 6 - 6 + n}{3}} = \frac{1}{6} \Rightarrow {(2)}^{\frac{n - 12}{3}} . {(3)}^{\frac{n - 12}{3}} = {(6)}^{- 1} \\ \Rightarrow {(6)}^{\frac{n - 12}{3}} = {(6)}^{- 1} \Rightarrow \frac{n - 12}{3} = - 1 \\ \Rightarrow n - 12 = - 3 \Rightarrow n = 12 - 3 = 9 \\ H e n c e, t h e r e q u i r e d v a l u e o f n i s 9 . \end{array}$

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4 months ago

Maths Maths NCERT Exemplar Solutions Class 11th Chapter Nine

A bag contains six white marbles and five red marbles. Find the number of ways in which four marbles can be drawn from the bag if:

(i) They can be of any colour.

(ii) Two must be white and two red.

(iii) They must all be of the same colour.

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Payal Gupta

Contributor-Level 10

This is a long answer type question as classified in NCERT Exemplar

$\begin{array}{l} T o t a l n u m b e r o f m a r b l e s = 6 w h i t e + 5 r e d = 11 m a r b l e s \\ (i) Since, w e h a v e t o d r a w 4 m a r b l e s o f a n y c o l o u r f r o m t h e 11 m a r b l e s \\ ∴ Required n u m b e r o f w a y s = {}_{11}{C_{4}} \\ (i i) I f 2 m u s t b e w h i t e a n d 2 m u s t b e r e d, t h e n t h e r e q u i r e d n u m b e r o f w a y s = {}_{6}{C_{2}} * {}_{5}{C_{2}} \\ (i i i) I f a l l t h e 4 m a r b l e s a r e o f t h e s a m e c o l o u r, t h e n t h e r e q u i r e d n u m b e r o f w a y s = {}_{6}{C_{4}} + {}_{5}{C_{4}} \\ H e n c e, t h e r e q u i r e d n u m b e r o f w a y s a r e \\ (i) {}_{11}{C_{4}} (i i) {}_{6}{C_{2}} * {}_{5}{C_{2}} (i i i) {}_{6}{C_{4}} + {}_{5}{C_{4}} \end{array}$

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4 months ago

A side of an equilateral triangle is 20 cm long. A second equilateral triangle is inscribed in it by joining the midpoints of the sides of the first triangle. The process is continued as shown in the accompanying diagram. Find the perimeter of the sixth inscribed equilateral triangle.

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Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} T h e s i d e o f t h e f i r s t e q u i l a t e r a l Δ A B C = 20 c m \\ B y j o i n i n g t h e m i d points o f t h e s i d e s o f t h i s t r i a n g l e, w e g e t t h e second e q u i l a t e r a l t r i a n g l e \\ w h i c h e a c h s i d e = \frac{20}{2} = 10 c m [\begin{array}{l} ? T h e l i n e j o i n i n g t h e m i d -points o f t w o s i d e s o f a t r i a n g l e \\ i s \frac{1}{2} a n d p a r a l l e l t o t h e t h i r d s i d e o f t h e t r i a n g l e \end{array}] \\ S i m i l a r l y, e a c h s i d e o f t h e t h i r d e q u i l a t e r a l t r i a n g l e = \frac{10}{2} = 5 c m \\ ∴ P e r i m e t e r o f f i r s t t r i a n g l e = 20 * 3 = 60 c m \\ P e r i m e t e r o f second t r i a n g l e = 10 * 3 = 30 c m \\ a n d t h e p e r i m e t e r o f t h i r d t r i a n g l e = 5 * 3 = 15 c m \\ T h e r e f o r e, t h e s e r i e s w i l l b e 60, 30, 15, \dots \\ w h i c h i s G . P . i n w h i c h a = 60, a n d r = \frac{30}{60} = \frac{1}{2} \\ N o w, w e h a v e t o f i n d t h e p e r i m e t e r o f s i x t h i n s c r i b e d e q u i l a t e r a l t r i a n g l e \\ ∴ a_{6} = a r^{6 - 1} \\ = 60 * {(\frac{1}{2})}^{5} = 60 * \frac{1}{32} = \frac{15}{8} c m \\ H e n c e, t h e r e q u i r e d p e r i m e t e r = \frac{15}{8} c m \end{array}$

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New answer posted

4 months ago

Maths Maths NCERT Exemplar Solutions Class 11th Chapter Nine

18 mice were placed in two experimental groups and one control group, with all groups equally large. In how many ways can the mice be placed into three groups?

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Payal Gupta

Contributor-Level 10

This is a long answer type question as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t 18 m i c e w e r e p l a c e d e q u a l l y i n t w o experimental g r o u p s a n d \\ o n e c o n t r o l g r o u p i . e . 3 g r o u p s \\ ∴ T h e r e q u i r e d n u m b e r o f a r r a n g e m e n t s = \frac{T o t a l a r r a n g e m e n t s}{E q u a l l y l i k e l y a r r a n g e m e n t s} \\ = \frac{18!}{6! 6! 6!} = \frac{18!}{{(6!)}^{3}} \\ H e n c e, t h e r e q u i r e d a r r a n g e m e n t s = \frac{18!}{{(6!)}^{3}} \end{array}$

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4 months ago

We know the sum of the interior angles of a triangle is . Show that the sums of the interior angles of polygons with 3, 4, 5, 6, . sides form an arithmetic progression. Find the sum of the interior angles for a 21-sided polygon.

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Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} Since, t h e s u m o f a l l interior a n g l e s o f a p o l y g o n o f n s i d e s = (2 n - 4) * 9 0^{0} \\ ∴ S u m o f interior a n g l e s o f a p o l y g o n o f 3 s i d e s = (2 * 3 - 4) * 9 0^{0} = 18 0^{0} \\ S u m o f interior a n g l e s o f a p o l y g o n o f 4 s i d e s = (2 * 4 - 4) * 9 0^{0} = 36 0^{0} \\ S i m i l a r l y, t h e s u m o f interior a n g l e s o f a p o l y g o n o f s i d e s 5, 6, 7, \dots a r e 54 0^{0}, 72 0^{0}, 90 0^{0}, \dots \\ T h e r e f o r e, t h e s e r i e s w i l l b e 18 0^{0}, 36 0^{0}, 54 0^{0}, 72 0^{0}, 90 0^{0}, \dots w h i c h i s A . P . \\ H e r e a = 18 0^{0}, d = 18 0^{0} \\ W e h a v e t o f i n d t h e s u m o f a l l interior a n g l e s o f a p o l y g o n o f 21 s i d e s i . e ., 19 t h t e r m \\ a_{n} = a + (n - 1) d \\ a_{19} = 18 0^{0} + (19 - 1) 18 0^{0} = 18 0^{0} + 18 * 18 0^{0} \\ = 18 0^{0} + 324 0^{0} = 342 0^{0} \\ H e n c e, t h e r e q u i r e d s u m o f interior a n g l e s = 342 0^{0} . \end{array}$

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4 months ago

Show that the middle term in the expansion of (x – 1/x)²ⁿis 1 * 3 *5 * … (2n -1)/n * (-2)ⁿ.

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Contributor-Level 10

This is a Long Answer Type Questions as classified in NCERT Exemplar
Sol:

$\begin{array}{l} G i v e n expression i s {(x - \frac{1}{x})}^{2 n} \\ N u m b e r o f t e r m s = 2 n + 1 = 9 (o d d) \\ ∴ M i d d l e t e r m {(\frac{n + 1}{2})}^{t h} t e r m = \frac{2 n + 1 + 1}{2} = (n + 1) t h t e r m . \\ G e n e r a l T e r m T_{r + 1} = {}_{n}{C_{r}} x^{n - r} y^{r} \\ ∴ T_{n + 1} = {}_{2 n}{C_{n}} {(x)}^{2 n - n} {(- \frac{1}{x})}^{n} = {}_{2 n}{C_{n}} {(x)}^{n} {(- 1)}^{n} . \frac{1}{x^{n}} \\ = {(- 1)}^{n} . {}_{2 n}{C_{n}} = {(- 1)}^{n} . \frac{2 n!}{n! (2 n - n)!} \\ = {(- 1)}^{n} . \frac{2 n!}{n! n!} = {(- 1)}^{n} . \frac{2 n (2 n - 1) (2 n - 2) (2 n - 3) \dots 1}{n! n (n - 1) (n - 2) (n - 3) \dots 1} \\ = {(- 1)}^{n} . \frac{2 n (2 n - 1) . 2 (n - 1) (2 n - 3) \dots 1}{n! n (n - 1) (n - 2) (n - 3) \dots 1} \\ = \frac{{(- 1)}^{n} . 2^{n} . [n (n - 1) (n - 1) \dots] . [(2 n - 1) . (2 n - 3) \dots 5.3.1]}{n! . n (n - 1) (n - 2) \dots 1} \\ = \frac{{(- 2)}^{n} [(2 n - 1) . (2 n - 3) \dots 5.3.1]}{n!} \\ = \frac{1 * 3 * 5 * \dots (2 n - 1)}{n!} * {(- 2)}^{n} \\ H e n c e, t h e m i d d l e t e r m = \frac{1 * 3 * 5 * \dots (2 n - 1)}{n!} * {(- 2)}^{n} \end{array}$

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4 months ago

If $p$ is a real number and if the middle term in the expansion of ${(\frac{p}{2} + 2)}^{8}$ is 1120, find $p$ .

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Maths Maths NCERT Exemplar Solutions Class 11th Chapter Nine

Contributor-Level 10

This is a Long Answer Type Questions as classified in NCERT Exemplar
Sol:

$\begin{array}{l} G i v e n expression i s {(\frac{P}{2} + 2)}^{8} \\ N u m b e r o f t e r m s = 8 + 1 = 9 (o d d) \\ ∴ M i d d l e t e r m {(\frac{n + 1}{2})}^{t h} t e r m = \frac{9 + 1}{2} = \frac{10}{2} = 5^{t h} t e r m \\ T_{5} = T_{4 + 1} = {}_{8}{C_{4}} {(\frac{P}{2})}^{8 - 4} {(2)}^{4} \\ = {}_{8}{C_{4}} \frac{P^{4}}{2^{4}} * 2^{4} = {}_{8}{C_{4}} P^{4} \\ N o w {}_{8}{C_{4}} P^{4} = 1120 \Rightarrow \frac{8 * 7 * 6 * 5}{4 * 3 * 2 * 1} . P^{4} = 1120 \\ \Rightarrow 70 P^{4} = 1120 \Rightarrow P^{4} = \frac{1120}{70} = 16 \\ \Rightarrow P^{4} = 2^{4} \Rightarrow P = \pm 2 \\ H e n c e, t h e r e q u i r e d v a l u e o f P = \pm 2 . \end{array}$

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4 months ago

A carpenter was hired to build 192 window frames. The first day he made five frames and each day, thereafter he made two more frames than he made the day before. How many days did it take him to finish the job?

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Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} H e r e, f i r s t t e r m a = 5 a n d t h e c o m m o n d i f f e r e n c e d = 2 l e t t h e c a r p e n t e r w i l l t a k e n d a y s \\ t o f i n i s h t h e j o b . \\ S_{n} = 192 \\ S_{n} = \frac{n}{2} [2 a + (n - 1) d] \\ 192 = \frac{n}{2} [2 * 5 + (n - 1) 2] \\ \Rightarrow 192 * 2 = n [10 + 2 n - 2] \Rightarrow 384 = n (2 n + 8) \\ \Rightarrow 384 = 2 n^{2} + 8 n \Rightarrow 2 n^{2} + 8 n - 384 = 0 \\ \Rightarrow n^{2} + 4 n - 192 = 0 \Rightarrow n^{2} + 16 n - 12 n - 192 = 0 \\ \Rightarrow n (n + 16) - 12 (n + 16) = 0 \Rightarrow (n - 12) (n + 16) = 0 \\ \Rightarrow n = 12 [? n \neq - 16] \\ H e n c e, t h e r e q u i r e d n u m b e r o f d a y s = 12 . \end{array}$

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New answer posted

4 months ago

Find the coefficient of $x^{4}$ in the expansion of ${(1 + x + x^{2} + x^{3})}^{11} .$

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Maths Maths NCERT Exemplar Solutions Class 11th Chapter Nine

Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar
Sol:

$\begin{array}{l} G i v e n expression i s {(1 + x + x^{2} + x^{3})}^{11} \\ = {[(1 + x) + x^{2} (1 + x)]}^{11} = {[(1 + x) (1 + x^{2})]}^{11} \\ = {(1 + x)}^{11} . {(1 + x^{2})}^{11} \\ E x p a n d i n g t h e a b o v e expression, w e g e t \\ ({}_{11}{C_{0}} + {}_{11}{C_{1}} x + {}_{11}{C_{2}} x^{2} + {}_{11}{C_{3}} x^{3} + {}_{11}{C_{4}} x^{4} + \dots) . ({}_{11}{C_{0}} + {}_{11}{C_{1}} x^{2} + {}_{11}{C_{2}} x^{4} + \dots) \\ = (1 + 11 x + 55 x^{2} + 165 x^{3} + 330 x^{4} + \dots) . (1 + 11 x^{2} + 55 x^{4} + \dots) \\ C o l l e c t i n g t h e t e r m s c o n t a i n i n g x^{4}, w e g e t \\ (55 + 605 + 330) x^{4} = 990 x^{4} \\ H e n c e, t h e c o e f f i c i e n t o f x^{4} = 990 \end{array}$

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4 months ago

If the $p$ -th and $q$ -th terms of a G.P. are $q$ and $p$ respectively, show that its $(p + q)$ -th term is ${(\frac{q}{p})}^{p - q}$

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