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B.J.B Autonomous College B.J.B Autonomous College Admission chances

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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9 months ago

I have passed the 10th grade and want to enroll for the 11th grade at B.J.B. Autonomous College. What is the process for the same?

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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Nine Exemplar Solution

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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Maths NCERT Exemplar Solutions Class 11th Chapter Seven Exemplar Solution

Contributor-Level 10

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

Sol:

$\begin{array}{l} L e t a b e t h e f i r s t t e r m a n d r b e t h e c o m m o n r a t i o o f a G . P . \\ G i v e n t h a t a_{p} = q \Rightarrow a r^{p - 1} = q \dots (i) \\ a n d a_{q} = p \Rightarrow a r^{q - 1} = p \dots (i i) \\ D i v i d i n g e q n . (i) b y e q n . (i i) w e g e t, \\ \frac{a r^{p - 1}}{a r^{q - 1}} = \frac{q}{p} \Rightarrow \frac{r^{p - 1}}{r^{q - 1}} = \frac{q}{p} \\ \Rightarrow r^{p - q} = \frac{q}{p} \Rightarrow r = {(\frac{q}{p})}^{\frac{1}{p - q}} \\ P u t t i n g t h e v a l u e o f r i n e q n . (i) w e g e t \\ a {(\frac{q}{p})}^{\frac{1}{p - q} \times p - 1} = q \\ a {(\frac{q}{p})}^{\frac{p - 1}{p - q}} = q \\ ∴ a = q . {(\frac{p}{q})}^{\frac{p - 1}{p - q}} \\ N o w T_{p + q} = a r^{p + q - 1} = q . {(\frac{p}{q})}^{\frac{p - 1}{p - q}} . {(\frac{q}{p})}^{\frac{1}{p - q} \times (p + q - 1)} \\ = q . {(\frac{p}{q})}^{\frac{p - 1}{p - q}} . {(\frac{q}{p})}^{\frac{p + q - 1}{p - q}} = q . {(\frac{p}{q})}^{\frac{p - 1}{p - q}} . {(\frac{p}{q})}^{\frac{- (p + q - 1)}{p - q}} \\ = q . {(\frac{p}{q})}^{\frac{p - 1}{p - q} - \frac{p + q - 1}{p - q}} = q . {(\frac{p}{q})}^{\frac{p - 1 - p - q + 1}{p - q}} \\ = q . {(\frac{p}{q})}^{\frac{- q}{p - q}} = q . {(\frac{q}{p})}^{\frac{q}{p - q}} = \frac{q^{\frac{q}{p - q} + 1}}{p^{\frac{q}{p - q}}} = \frac{q^{\frac{p}{p - q}}}{p^{\frac{q}{p - q}}} = {[\frac{q^{p}}{p^{q}}]}^{\frac{1}{p - q}} \\ H e n c e, t h e r e q u i r e d t e r m = {[\frac{q^{p}}{p^{q}}]}^{\frac{1}{p - q}} . \end{array}$

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9 months ago

A convex polygon has 44 diagonals. Find the number of its sides.

[Hint: Polygon of n sides has (ⁿC_{2 – n}) number of diagonals.]

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Maths NCERT Exemplar Solutions Class 11th Chapter Eight Exemplar Solution

Contributor-Level 10

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

$\begin{array}{l} G i v e n t h a t q u e s t i o n n u m b e r 1 a n d 2 a r e c o m p u l s o r y . \\ ∴ T h e r e m a i n i n g q u e s t i o n s a r e 5 - 2 = 3 \\ L e t n b e t h e n u m b e r o f s i d e s i n a p o l y g o n . \\ Since, P o l y g o n o f n s i d e s h a s ({}_{n}{C_{2}} - n) n u m b e r o f d i a g o n a l s \\ ∴ {}_{n}{C_{2}} - n = 44 = \frac{n!}{2! (n - 2)!} - n = 44 \\ \Rightarrow \frac{n (n - 1) (n - 2)!}{2 . (n - 2)!} - n = 44 \Rightarrow \frac{n (n - 1)}{2} - n = 44 \\ \Rightarrow \frac{n^{2} - n - 2 n}{2} = 44 \Rightarrow n^{2} - 3 n = 88 \Rightarrow n^{2} - 3 n - 88 = 0 \\ \Rightarrow n^{2} - 11 n + 8 n - 88 = 0 \Rightarrow n (n - 11) + 8 (n - 11) = 0 \\ \Rightarrow (n - 11) (n + 8) = 0 ∴ n = 11 a n d n = - 8 [? n \neq - 8] \\ S o, n = 11 \\ H e n c e, t h e r e q u i r e d n u m b e r o f s i d e s = 11 . \end{array}$

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9 months ago

If the coefficient of second, third, and fourth terms in the expansion of ${(1 + x)}^{2 n}$ are in A.P. Show that $2 n^{2} - 9 n + 7 = 0 .$

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Maths NCERT Exemplar Solutions Class 11th Chapter Seven Exemplar Solution

Contributor-Level 10

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

$\begin{array}{l} G i v e n expression i s {(1 + x)}^{2 n} \\ H e r e, c o e f f i c i e n t o f 2 n d t e r m = {}_{2 n}{C_{1}} \\ C o e f f i c i e n t o f 3 r d t e r m = {}_{2 n}{C_{2}} \\ a n d c o e f f i c i e n t o f 4 t h t e r m = {}_{2 n}{C_{3}} \\ G i v e n t h a t {}_{2 n}{C_{1}}, {}_{2 n}{C_{2}} a n d {}_{2 n}{C_{3}} a r e i n A . P \\ ∴ {}_{2 n}{C_{2}} - {}_{2 n}{C_{1}} = {}_{2 n}{C_{3}} - {}_{2 n}{C_{2}} \\ \Rightarrow 2 . {}_{2 n}{C_{2}} = {}_{2 n}{C_{1}} + {}_{2 n}{C_{3}} \\ \Rightarrow 2 . \frac{2 n!}{2! (2 n - 2)!} = \frac{2 n!}{(2 n - 1)!} + \frac{2 n!}{3! (2 n - 3)!} \\ \Rightarrow 2 [\frac{2 n (2 n - 1) (2 n - 2)!}{2 \times 1 \times (2 n - 2)!}] = \frac{2 n (2 n - 1)!}{(2 n - 1)!} + \frac{2 n (2 n - 1) (2 n - 2) (2 n - 3)!}{3 \times 2 \times 1 \times (2 n - 3)!} \\ \Rightarrow n (2 n - 1) = n + \frac{n (2 n - 1) (2 n - 2)}{6} \\ \Rightarrow 2 n - 1 = 1 + \frac{(2 n - 1) (2 n - 2)}{6} \\ \Rightarrow 12 n - 6 = 6 + 4 n^{2} - 4 n - 2 n + 2 \\ \Rightarrow 12 n - 12 = 4 n^{2} - 6 n + 2 \\ \Rightarrow 4 n^{2} - 6 n - 12 n + 2 + 12 = 0 \\ \Rightarrow 4 n^{2} - 18 n + 14 = 0 \\ \Rightarrow 2 n^{2} - 9 n + 7 = 0 \\ H e n c e p r o v e d . \end{array}$

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9 months ago

In an examination, a student has to answer 4 questions out of 5 questions; questions 1 and 2 are however compulsory. Determine the number of ways in which the student can make the choice.

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Maths NCERT Exemplar Solutions Class 11th Chapter Nine Exemplar Solution

Contributor-Level 10

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

$\begin{array}{l} G i v e n t h a t q u e s t i o n n u m b e r 1 a n d 2 a r e c o m p u l s o r y . \\ ∴ T h e r e m a i n i n g q u e s t i o n s a r e 5 - 2 = 3 \\ T o t a l n u m b e r o f q u e s t i o n s t o b e a t t e m p t e d = 4 q u e s t i o n s 1 a n d 2 a r e c o m p u l s o r y . \\ S o o n l y 2 q u e s t i o n s a r e t o b e d o n e o u t o f 3 q u e s t i o n s \\ T h e r e f o r e n u m b e r o f w a y s = {}_{3}{C_{2}} = {}_{3}{C_{3 - 2}} = {}_{3}{C_{1}} = 3 [? {}_{n}{C_{r}} = {}_{n}{C_{n - r}}] \\ H e n c e, t h e r e q u i r e d n u m b e r o f w a y s = 3 . \end{array}$

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9 months ago

A man accepts a position with an initial salary of Rs 5200 per month. It is understood that he will receive an automatic increase of Rs 320 in the very next month and each month thereafter.

(a) Find his salary for the tenth month.

(b) What is his total earnings during the first year?

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Maths NCERT Exemplar Solutions Class 11th Chapter Seven Exemplar Solution

Contributor-Level 10

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

Sol:

$\begin{array}{l} G i v e n t h a t f i x e d i n c r e m e n t i n t h e s a l a r y o f a m a n = 320 e a c h m o n t h \\ I n i t i a l s a l a r y = 5200 w h i c h m a k e s a n A . P . \\ w h o s e f i r s t t e r m (a) = 5200 a n d c o m m o n d i f f e r e n c e (d) = 320 \\ (i) S a l a r y f o r t h e t e n t h m o n t h \\ a_{10} = a + (n - 1) d \\ = 5200 + (10 - 1) \times 320 = 5200 + 2880 = 8080 \\ (i i) T o t a l e a r n i n g d u r i n g t h e f i r s t y e a r (12 m o n t h s) \\ S_{12} = \frac{12}{2} [2 \times 5200 + (12 - 1) \times 320] [? S_{n} = \frac{n}{2} [2 a + (n - 1) d]] \\ = 6 [10400 + 3520] = 6 \times 13920 = 83520 \\ H e n c e, t h e r e q u i r e d a m o u n t i s (i) 8080 a n d (i i) 83520 \end{array}$

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9 months ago

In a certain city, all telephone numbers have six digits, the first two digits always being 41 or 42 or 46 or 62 or 64. How many telephone numbers have all six digits distinct?

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Maths NCERT Exemplar Solutions Class 11th Chapter Eight Exemplar Solution

Contributor-Level 10

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

$\begin{array}{l} I f f i r s t t w o d i g i t s i s 41, t h e n t h e r e m a i n i n g 4 d i g i t s c a n b e a r r a n g e d i n {}_{8}{P_{4}} w a y s \\ = \frac{8!}{(8 - 4)!} = \frac{8!}{4!} = \frac{8 \times 7 \times 6 \times 5 \times 4!}{4!} = 1680 \\ S i m i l a r l y f i r s t t w o d i g i t s c a n b e 42 o r 46 o r 62 o r 64 . \\ ∴ T o t a l n u m b e r o f t e l e p h o n e n u m b e r s h a v e a l l d i g i t s d i s t i n c t = 5 \times 1680 = 8400 \\ H e n c e, t h e r e q u i r e d t e l e p h o n e n u m b e r s = 8400 . \end{array}$

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9 months ago

Find the value of $r$ , if the coefficients of the (2r + 4)^th and ^th terms in the expansion of ${(1 + x)}^{18}$ are equal.

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DE-CODE Preparation Exam Application Form

Contributor-Level 10

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

$\begin{array}{l} G e n e r a l T e r m T_{r + 1} = {}_{n}{C_{r}} x^{n - r} y^{r} \\ F o r c o e f f i c i e n t s o f {(2 r + 4)}^{t h} t e r m, w e h a v e \\ T_{2 r + 4} = T_{2 r + 3 + 1} = {}_{18}{C_{2 r + 3}} {(1)}^{18 - 2 r - 3} . x^{2 r + 3} \\ ∴ C o e f f i c i e n t s o f {(2 r + 4)}^{t h} t e r m = {}_{18}{C_{2 r + 3}} \\ S i m i l a r l y, T_{r - 2} = T_{r - 3 + 1} = {}_{18}{C_{r - 3}} {(1)}^{18 - r + 3} . x^{r - 3} \\ ∴ C o e f f i c i e n t s o f {(r - 2)}^{t h} t e r m = {}_{18}{C_{r - 3}} \\ A s p e r t h e c o n d i t i o n o f t h e q u e s t i o n s, w e h a v e {}_{18}{C_{2 r + 3}} = {}_{18}{C_{r - 3}} \\ \Rightarrow 2 r + 3 + r - 3 = 18 \Rightarrow 3 r = 18 \Rightarrow r = 6 \end{array}$

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9 months ago

Where can I access the DE-CODE entrance exam application form?

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Sayeba Naushad

Contributor-Level 10

You can access the DE-CODE application form on the official website of the institute conducting the DECODE exam. Make sure to register using a valid email ID and mobile number.

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9 months ago

Architecture Entrance Exam Exam

Can I apply for NATA without completing Class 12?

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Aishwarya Garg

Contributor-Level 7

Yes, you can apply for NATA without completing Class 12, provided you are appearing for the Class 12 board exams (or 10+3 Diploma final exams) in the same academic year.

Eligibility Clarification:
Candidates appearing for the 10+2 exam with Physics, Chemistry, and Mathematics
OR

Candidates appearing for the 10+3 Diploma exam with Mathematics as a subject are eligible to apply for NATA provisionally.
However, you must pass the qualifying exam with the required subjects and marks before admission to a B.Arch. course.

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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Seven Exemplar Solution

If 20 lines are drawn in a plane such that no two of them are parallel and no three are concurrent, in how many points will they intersect each other?

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Indian Institute for Aeronautical Engineering & Information Technology (IIAEIT) Colleges in Pune

Contributor-Level 10

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

$\begin{array}{l} G i v e n t h a t o u t o f 20 l i n e s, n o t w o l i n e s a r e p a r a l l e l a n d n o t h r e e l i n e s a r e c o n c u r r e n t . \\ T h e r e f o r e, n u m b e r o f point o f intersection \\ = {}_{20}{C_{2}} [? f o r a n y point o f intersection, w e n e e d t w o l i n e s] \\ = \frac{20.19}{2.1} = 190 \\ H e n c e, t h e r e q u i r e d n u m b e r o f points = 190 . \end{array}$

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9 months ago

Is studying BTech at IIAEIT expensive?

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Sejal Baveja

Contributor-Level 10

The cost of studying BTech at IIAEIT includes tuition fees, hostel charges, lab and exam fees, and other administrative expenses. The total tuition fees for BTech is INR 9.2 lakh. However, candidates can also avail scholarships through various options such as AERO-CET scores, defence and sports quota.

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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Nine Exemplar Solution

A man saved Rs 66000 in 20 years. In each succeeding year after the first year, he saved Rs 200 more than what he saved in the previous year. How much did he save in the first year?

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Maths NCERT Exemplar Solutions Class 11th Chapter Seven Exemplar Solution

Contributor-Level 10

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

Sol:

$\begin{array}{l} L e t x b e s a v e d i n f i r s t y e a r . \\ A n n u a l i n c r e m e n t = 200 \\ w h i c h f o r m s a n A . P . \\ f i r s t t e r m = a a n d c o m m o n d i f f e r e n c e d = 200 \\ n = 20 y e a r s \\ ∴ S_{n} = \frac{n}{2} [2 a + (n - 1) d] = S_{20} = \frac{20}{2} [2 a + (20 - 1) 200] \\ \Rightarrow 66000 = 10 [2 a + 3800] \Rightarrow 6600 = 2 a + 3800 \\ \Rightarrow 2 a = 6600 - 3800 \Rightarrow 2 a = 2800 \Rightarrow a = 1400 \\ H e n c e, t h e m a n s a v e d 1400 i n t h e f i r s t y e a r . \end{array}$

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9 months ago

Find the number of integers greater than 7000 that can be formed with the digits 3, 5, 7, 8, and 9 where no digits are repeated.

[Hint: Besides 4-digit integers greater than 7000, five-digit integers are always greater than 7000.]

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Maths NCERT Exemplar Solutions Class 11th Chapter Eight Exemplar Solution

Contributor-Level 10

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

$\begin{array}{l} G i v e n t h a t a l l t h e 5 d i g i t n u m b e r s a r e g r e a t e r t h a n 7000 . \\ S o, t h e w a y s o f f o r m a t i n g 5 - d i g i t n u m b e r s = 5 \times 4 \times 3 \times 2 \times 1 = 120 \\ N o w a l l t h e f o u r d i g i t n u m b e r g r e a t e r t h a n 7000 c a n b e f o r m e d a s f o l l o w s . \\ T h o u s a n d p l a c e c a n b e f i l l e d w i t h 3 w a y s \\ H u n d r e d p l a c e c a n b e f i l l e d w i t h 4 w a y s \\ T e n t h s p l a c e c a n b e f i l l e d w i t h 3 w a y s \\ U n i t s p l a c e c a n b e f i l l e d w i t h 2 w a y s \\ S o, t h e t o t a l n u m b e r o f 4 - d i g i t n u m b e r s = 3 \times 4 \times 3 \times 2 = 72 \\ ∴ T o t a l n u m b e r o f integers = 120 + 72 = 192 \\ H e n c e, t h e r e q u i r e d n u m b e r o f integers = 192 \end{array}$

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9 months ago

Find the sixth term of the expansion ${(y^{\frac{1}{2}} + x^{\frac{1}{3}})}^{n}$ , if the binomial coefficient of the third term from the end is 45.

[Hint: Binomial coefficient of third term from the end = Binomial coefficient of third term from the beginning = ⁿC₂.]

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Teaching & Education UGC NET Application Form

Contributor-Level 10

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

$\begin{array}{l} T h e g i v e n expression i s {(y^{\frac{1}{2}} + x^{\frac{1}{3}})}^{n}, since t h e b i n o m i a l c o e f f i c i e n t o f t h i r d \\ t e r m f r o m t h e e n d = b i n o m i a l c o e f f i c i e n t o f t h i r d t e r m f r o m t h e b e g i n n i n g = {}_{n}{C_{2}} \\ ∴ {}_{n}{C_{2}} = 45 \\ \Rightarrow \frac{n (n - 1)}{2} = 45 \Rightarrow n^{2} - n = 90 \\ \Rightarrow n^{2} - n - 90 = 0 \Rightarrow n^{2} - 10 n + 9 n - 90 = 0 \\ \Rightarrow n (n - 10) + 9 (n - 10) = 0 \Rightarrow (n - 10) (n + 9) = 0 \\ \Rightarrow n = 10, n = - 9 \Rightarrow n = 10, n \neq - 9 \\ S o, t h e g i v e n expression b e c o m e s {(y^{\frac{1}{2}} + x^{\frac{1}{3}})}^{10} \\ S i x t h t e r m i n t h i s expression \\ T_{6} = T_{5 + 1} = {}_{10}{C_{5}} {(y^{\frac{1}{2}})}^{10 - 5} {(x^{\frac{1}{3}})}^{5} = {}_{10}{C_{5}} y^{\frac{5}{2}} . x^{\frac{5}{3}} \\ = 252 y^{\frac{5}{2}} . x^{\frac{5}{3}} \\ H e n c e, t h e r e q u i r e d t e r m = 252 y^{\frac{5}{2}} . x^{\frac{5}{3}} \end{array}$

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9 months ago

How to apply for the UGC NET 2026 June session?

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Shikha Shukla

Contributor-Level 7

Candidates can check below the process to apply for the NTA UGC NET application form for the June session Visit at- ugcnet.nta.nic.in, Click on the 'Application form' link, mentioned at the bottom of the home page, Click on the 'New Registration', Enter the required details and complete the registration, Fill out the complete online form, Upload scanned pictures of passport size photo (10-200 kb) and signature (10-50 kb). Both images should be in JPEG format, Pay the application fee (Gen - INR 1150; OBC/EWS - INR 600; SC/ST/PwD/ Transgender - INR 325), Save and submit the form, Download the confirmation page.

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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Seven Exemplar Solution

If ⁿC_{r – 1} = 36, ⁿC_r = 84, and ⁿC_{r + 1}= 126, then find ^rC₂.

[Hint: Form equation using C(n, r-1), C(n, r), and C(n, r+1) to find the value of r.]

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Maths NCERT Exemplar Solutions Class 11th Chapter Nine Exemplar Solution

Contributor-Level 10

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

$\begin{array}{l} G i v e n t h a t {}_{n}{C_{r - 1}} = 36 \dots (i) \\ {}_{n}{C_{r}} = 84 \dots (i i) \\ {}_{n}{C_{r + 1}} = 126 \dots (i i i) \\ D i v i d i n g e q n . (i) b y e q n . (i i) w e g e t \\ \frac{{}_{n}{C_{r - 1}}}{{}_{n}{C_{r}}} = \frac{36}{84} \\ \Rightarrow \frac{\frac{n!}{(r - 1)! (n - r + 1)!}}{\frac{n!}{r! (n - r)!}} = \frac{3}{7} [? {}_{n}{C_{r}} = \frac{n!}{r! (n - r)!}] \\ \Rightarrow \frac{n!}{(r - 1)! (n - r + 1)!} \times \frac{r! (n - r)!}{n!} = \frac{3}{7} \\ \Rightarrow \frac{r . (r - 1)! (n - r)!}{(r - 1)! (n - r + 1)! + (n - r)!} = \frac{3}{7} \\ \Rightarrow \frac{r}{n - r + 1} = \frac{3}{7} \Rightarrow 3 n - 3 r + 3 = 7 r \\ \Rightarrow 3 n - 10 r = - 3 \dots (i v) \\ D i v i d i n g e q n . (i i) b y e q n . (i i i) w e g e t \\ \frac{{}_{n}{C_{r}}}{{}_{n}{C_{r + 1}}} = \frac{84}{126} \Rightarrow \frac{\frac{n!}{r! (n - r)!}}{\frac{n!}{(r + 1)! (n - r - 1)!}} = \frac{2}{3} \\ \Rightarrow &th \end{array}$

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9 months ago

The first term of an A.P. is $a$ , and the sum of the first $p$ terms is zero, show that the

sum of its next $q$ terms is $\frac{- a (p + q) q}{p - 1}$ .

[Hint: Required sum = $S_{p + q} - S_{p}$ ]

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Maths NCERT Exemplar Solutions Class 11th Chapter Eight Exemplar Solution

Contributor-Level 10

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

Sol:

$\begin{array}{l} G i v e n t h a t a_{1} = a a n d S_{p} \\ S u m o f n e x t q t e r m s o f t h e g i v e n A . P . = S_{p + q} - S_{p} \\ ∴ S_{p + q} = \frac{p + q}{2} [2 a + (p + q - 1) d] \\ a n d S_{p} = \frac{p}{2} [2 a + (p - 1) d] = 0 \\ \Rightarrow 2 a + (p - 1) d = 0 \Rightarrow (p - 1) d = - 2 a \\ \Rightarrow d = \frac{- 2 a}{p - 1} \\ S u m o f n e x t q t e r m s = S_{p + q} - S_{p} \\ = \frac{p + q}{2} [2 a + (p + q - 1) d] - 0 \\ = \frac{p + q}{2} [2 a + (p + q - 1) (\frac{- 2 a}{p - 1})] \\ = \frac{p + q}{2} [2 a + \frac{(p - 1) (- 2 a)}{p - 1} - \frac{2 a q}{p - 1}] \\ = \frac{p + q}{2} [2 a - 2 a - \frac{2 a q}{p - 1}] = \frac{p + q}{2} (- \frac{2 a q}{p - 1}) \\ = \frac{- a (p + q) q}{p - 1} \\ H e n c e, t h e r e q u i r e d s u m = \frac{- a (p + q) q}{p - 1} \end{array}$

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9 months ago

Find the coefficient of $\frac{1}{x^{17}}$ in the expansion of x⁴– 1/x^{3 15}

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