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New answer posted•a year ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Nine

The third term of a G.P. is 4. The product of its first 5 terms is

(a) $4^{3}$

(b) $4^{4}$

(c) $4^{5}$

(d) None of these

3 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t T_{3} = 4 \\ \Rightarrow a r^{3 - 1} = 4 [? T_{n} = a r^{n - 1}] \\ \Rightarrow a r^{2} = 4 \\ P r o d u c t o f f i r s t 5 t e r m s = a . a r . a r^{2} . a r^{3} . a r^{4} = a^{5} r^{10} = {(a r^{2})}^{5} = {(4)}^{5} \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

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New answer posted•a year ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Nine

If the sum of terms of an A.P. is given by , then the common difference of the A.P. is

(a) 3

(b) 2

(c) 6

(d) 4

2 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t S_{n} = 3 n + 2 n^{2} \\ S_{1} = 3 (1) + 2 {(1)}^{2} = 5 \\ S_{2} = 3 (2) + 2 {(2)}^{2} = 14 \\ S_{1} = a_{1} = 5 \\ S_{2} - S_{1} = a_{2} = 14 - 5 = 9 \\ ∴ C o m m o n d i f f e r e n c e d = a_{2} - a_{1} = 9 - 5 = 4 \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

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New answer posted•a year ago

Exemplar Solution Class 11th

Given 5 different green dyes, four different blue dyes, and three different red dyes, the number of combinations of dyes that can be chosen, taking at least one green and one blue dye, is:

(a) 3600

(b) 3720

(c) 3800

(d) 3600

[Hint: Possible numbers of choosing or not choosing 5 green dyes, 4 blue dyes, and 3 red dyes are $2^{5}, 2^{4}, 2^{3}$ , respectively.]

7 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} P o s s i b l e n u m b e r o f choosing 5 d i f f e r e n t g r e e n d y e s = 2^{5} \\ P o s s i b l e n u m b e r o f choosing 4 b l u e d y e s = 2^{4} \\ a n d p o s s i b l e n u m b e r o f choosing 3 r e d d y e s = 2^{3} \\ I f a t l e a s t o n e b l u e a n d o n e g r e e n d y e s a r e s e l e c t e d t h e n t h e t o t a l n u m b e r o f s e l e c t i o n \\ = (2^{5} - 1) \times (2^{4} - 1) \times 2^{3} = 31 \times 15 \times 8 = 3720 \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

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New answer posted•a year ago

Exemplar Solution Class 11th

The number of words that can be formed out of the letters of the word "ARTICLE" so that vowels occupy the even places is:

(a) 1440

(b) 144

(c) 7!

(d) ⁴C₄ × ³C₃

4 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} T o t a l n u m b e r o f l e t t e r s i n t h e' A R T I C L E' i s 7 o u t w h i c h A, E, I a r e v o w e l s a n d \\ R, T, C, L a r e c o n s o n a n t s \\ G i v e n t h a t v o w e l s o c c u p y e v e n p l a c e \\ ∴ P o s s i b l e a r r a n g e m e n t c a n b e s h o w n a s b e l o w \\ C, V, C, V, C, V, C i . e . o n 2 n d, 4 t h a n d 6 t h p l a c e s \\ T h e r e f o r e, n u m b e r o f a r r a n g e m e n t = {}_{3}{P_{3}} = 3! = 6 w a y s \\ N o w c o n s o n a n t s c a n b e p l a c e d a t 1, 3, 5 a n d 7 t h p l a c e \\ ∴ N u m b e r o f a r r a n g e m e n t = {}_{4}{P_{4}} = 4! = 24 \\ S o, t h e t o t a l n u m b e r o f a r r a n g e m e n t s = 6 \times 24 = 144 \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

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New answer posted•a year ago

Exemplar Solution Class 11th

The total number of 9-digit numbers which have all different digits is:

(a) 10!

(b) 9!

(c) 9 × 9!

(d) 10 × 10!

6 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} W e h a v e t o f o r m 9 d i g i t n u m b e r s f r o m 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 a n d w e k n o w t h a t 0 c a n n o t b e p u t \\ o n e x t r e m e l y l e f t p l a c e . \\ S o, f i r s t p l a c e f r o m t h e l e f t s i d e c a n b e f i l l e d i n 9 w a y s . \\ N o w r e p e t i t i o n i s n o t a l l o w e d . S o, t h e r e m a i n i n g 8 p l a c e s c a n b e f i l l e d i n 9! \\ ∴ T h e r e q u i r e d n u m b e r o f w a y s = 9 \times 9! \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

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New answer posted•a year ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Nine

If $p$ -th, $q$ -th, and $r$ -th terms of an A.P. and G.P. are both $a, b, c$ respectively, show that $\frac{a}{b - c} \cdot \frac{b}{c - a} \cdot \frac{c}{a - b} = 1 .$

3 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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

Sol:

$\begin{array}{l} L e t A b e t h e f i r s t t e r m a n d d b e t h e c o m m o n d i f f e r e n c e o f t h e g i v e n A . P . a n d x a n d R b e t h e \\ f i r s t t e r m a n d c o m m o n r a t i o r e s p e c t i v e l y o f t h e G . P . \\ ∴ A + (p - 1) d = a \dots (i) \\ A + (q - 1) d = b \dots (i i) \\ a n d A + (r - 1) d = c \dots (i i i) \\ F o r G . P ., w e h a v e \\ x R^{p - 1} = a \dots (i v) \\ x R^{q - 1} = b \dots (v) \\ a n d x R^{r - 1} = c \dots (v i) \\ S u b t r a c t i n g e q n . (i i) f r o m e q n . (i) w e g e t \\ (p - q) d = a - b \dots (v i i) \\ S i m i l a r l y, (q - r) d = b - c \dots (v i i i) \\ a n d (r - p) d = c - a \dots (i x) \\ N o w w e h a v e t o p r o v e t h a t \\ a^{b - c} . b^{c - a} . c^{a - b} = 1 \\ L . H . S . = a^{b - c} . b^{c - a} . c^{a - b} \\ = {[x R^{p - 1}]}^{(q - r) d} . {[x R^{q - 1}]}^{(r - p) d} . {[x R^{r - 1}]}^{(p - q) d} [\begin{array}{l} F r o m e q . (i), (i i), (i i i), (i v), (v), (v i), \\ (v i i), (v i i i), (i x) \end{array}] \\ = x^{(q - r) d} . R^{(p - 1) (q - r) d} . x^{(r - p) d} . R^{(q - 1) (r - p) d} . x^{(p - q) d} . R^{(r - 1) (p - q) d} \\ = x^{(q - r) d + (r - p) d + (p - q) d} . R^{(p - 1) (q - r) d + (q - 1) (r - p) d + (r - 1) (p - q) d} \\ = x^{(q - r + r - p + p - q) d} . R^{(p q - p r - q + r + q r - p q - r + p + p r - q r - p + q) d} \\ = x^{(0) d} . R^{(0) d} = x^{0} . R^{0} = 1 R . H . S . \\ L . H . S . = R . H . S . H e n c e p r o v e d . \end{array}$

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New answer posted•a year ago

Exemplar Solution Class 11th

The number of terms in the expansion of ${(a + b)}^{n}$ where $n \in ℕ$ is one less than the power $n$ .

6 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

This is a True or False Type Questions as classified in NCERT Exemplar
Sol:

$\begin{array}{l} Since, t h e n u m b e r o f t e r m s i n t h e g i v e n expression i s {(a + b)}^{n} i s 1 m o r e t h a n n i . e ., n + 1 \\ H e n c e, t h e g i v e n s t a t e m e n t i s' F a l s e' . \end{array}$

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New answer posted•a year ago

Exemplar Solution Class 11th

If the expansion of ${(x - \frac{1}{x^{2}})}^{2 n}$ contains a term independent of $x$ , then $n$ is a multiple of 2.

4 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

This is a True or False Type Questions as classified in NCERT Exemplar
Sol:

$\begin{array}{l} G i v e n expression i s {(x - \frac{1}{x^{2}})}^{2 n} \\ T_{r + 1} = {}_{2 n}{C_{r}} {(x)}^{2 n - r} {(- \frac{1}{x^{2}})}^{r} = {}_{2 n}{C_{r}} {(x)}^{2 n - r} {(- 1)}^{r} . \frac{1}{x^{2 r}} \\ = {}_{2 n}{C_{r}} {(x)}^{2 n - r - 2 r} {(- 1)}^{r} = {}_{2 n}{C_{r}} {(x)}^{2 n - 3 r} {(- 1)}^{r} \\ F o r t h e t e r m i n d e p e n d e n t o f x, 2 n - 3 r = 0 \\ ∴ r = \frac{2 n}{3} w h i c h i s n o t a n integer a n d t h e expression i s n o t p o s s i b l e t o b e t r u e . \\ H e n c e, t h e g i v e n s t a t e m e n t i s' F a l s e' . \end{array}$

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New answer posted•a year ago

Exemplar Solution Class 11th

The number of ways in which we can choose a committee from four men and six women so that the committee includes at least two men and exactly twice as many women as men is:

(a) 94

(b) 126

(c) 128

(d) None

7 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} N u m b e r o f m e n = 4 \\ N u m b e r o f w o m e n = 6 \\ W e a r e g i v e n t h a t t h e c o m m i t t e e i n c l u d e s 2 m e n a n d e x a c t l y t w i c e a s m a n y w o m e n a s m e n . \\ T h u s, t h e p o s s i b l e s e l e c t i o n c a n b e \\ 2 m e n a n d 4 w o m e n a n d 3 m e n a n d 6 w o m e n . \\ S o, t h e n u m b e r o f c o m m i t t e e = {}_{4}{C_{2}} \times {}_{6}{C_{4}} + {}_{4}{C_{3}} \times {}_{6}{C_{6}} \\ = 6 \times 15 + 4 \times 1 = 90 + 4 = 94 \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

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New answer posted•a year ago

Exemplar Solution Class 11th

The last two digits of the number $3^{400}$ are 01.

4 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

This is a True or False Type Questions as classified in NCERT Exemplar
Sol:

$\begin{array}{l} G i v e n t h a t 3^{400} = {(9)}^{200} = {(10 - 1)}^{200} \\ ∴ {(10 - 1)}^{200} = {}_{200}{C_{0}} {(10)}^{200} + {}_{200}{C_{1}} {(10)}^{199} + \dots - {}_{200}{C_{199}} {(10)}^{1} + {}_{200}{C_{200}} {(1)}^{200} \\ = 1 0^{200} - 200 \times 1 0^{199} + \dots - 10 \times 200 + 1 \\ S o, i t i s c l e a r t h a t l a s t t w o d i g i t s a r e 01 . \\ H e n c e, t h e g i v e n s t a t e m e n t i s' T r u e' . \end{array}$

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