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New answer posted

a year ago

Class 11th Maths NCERT Exemplar Solutions Class 11th Chapter Nine

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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alok kumar singh

Contributor-Level 10

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

Sol:

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

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New answer posted

a year ago

Class 11th Exemplar Solution

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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Payal Gupta

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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New answer posted

a year ago

Class 11th Exemplar Solution

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

0 Follower 5 Views

Vishal Baghel

Contributor-Level 10

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

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

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New answer posted

a year ago

Class 11th Exemplar Solution

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.

0 Follower 2 Views

Payal Gupta

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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New answer posted

a year ago

Class 11th Maths NCERT Exemplar Solutions Class 11th Chapter Nine

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?

0 Follower 15 Views

alok kumar singh

Contributor-Level 10

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

Sol:

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

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New answer posted

a year ago

Class 11th Exemplar Solution

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?

0 Follower 4 Views

Payal Gupta

Contributor-Level 10

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

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

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New answer posted

a year ago

Class 11th Exemplar Solution

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.

0 Follower 6 Views

Vishal Baghel

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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New answer posted

a year ago

Class 11th 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?

0 Follower 2 Views

Payal Gupta

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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New answer posted

a year ago

Class 11th Maths NCERT Exemplar Solutions Class 11th Chapter Nine

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?

0 Follower 3 Views

alok kumar singh

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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New answer posted

a year ago

Class 11th Exemplar Solution

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.]

0 Follower 3 Views

Payal Gupta

Contributor-Level 10

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

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

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