Page 52 - Exemplar Solution - Updates, Questions & Discussions on Exemplar Solution at Shiksha

Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Nine

If the sum of $p$ terms of an A.P. is $q$ and the sum of $q$ terms is $p$ , show that the sum of $p + q$ terms is $- (p + q)$ . Also, find the sum of the first $p - q$ terms ( $p > q$ ).

7 Views|Posted a year ago

Asked by Alok Kumar Singh

S

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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

Sol:

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

Upvote

Exemplar Solution Class 11th

The sum of coefficients of the two middle terms in the expansion of ${(1 + x)}^{2 n - 1}$ is equal to ${}_{2 n - 1}{C_{n}}$ .

6 Views|Posted a year ago

Asked by Vishal Baghel

S

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

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

$\begin{array}{l} G i v e n expression i s {(1 + x)}^{2 n - 1} \\ N u m b e r o f t e r m s = 2 n - 1 + 1 = 2 n (e v e n) \\ ∴ M i d d l e t e r m s a r e \frac{2 n}{2} t h t e r m a n d {(\frac{2 n}{2} + 1)}^{t h} t e r m s = n t h t e r m s a n d (n + 1) t h t e r m s \\ C o e f f i c i e n t o f n t h t e r m = {}_{2 n - 1}{C_{n - 1}} \\ a n d t h e c o e f f i c i e n t o f (n + 1) t h t e r m = {}_{2 n - 1}{C_{n}} \\ S u m o f t h e c o e f f i c i e n t s = {}_{2 n - 1}{C_{n - 1}} + {}_{2 n - 1}{C_{n}} \\ = {}_{2 n - 1 + 1}{C_{n}} = {}_{2 n}{C_{n}} [? {}_{n}{C_{r}} + {}_{n}{C_{r - 1}} = {}_{n + 1}{C_{r}}] \\ H e n c e, t h e g i v e n s t a t e m e n t i s' F a l s e' . \end{array}$

Upvote

Exemplar Solution Class 11th

The number of 5-digit telephone numbers having at least one of their digits repeated is:

(a) 90,000

(b) 10,000

(c) 30,240

(d) 69,760

8 Views|Posted a year ago

Asked by Payal Gupta

S

Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} T o t a l n u m b e r o f 5 - d i g i t t e l e p h o n e n u m b e r i f a l l t h e d i g i t s a r e r e p e a t e d = {(10)}^{5} \\ [? d i g i t s a r e f r o m 0 t o 9] \\ I f d i g i t s a r e n o t r e p e a t e d, t h e n 5 - d i g i t t e l e p h o n e s, c a n b e f o r m e d i n {}_{10}{P_{5}} w a y s \\ ∴ Required n u m b e r o f w a y s = {(10)}^{5} - {}_{10}{P_{5}} \\ = 100000 - \frac{10!}{(10 - 5)!} = 100000 - \frac{10 \times 9 \times 8 \times 7 \times 6 \times 5!}{5!} \\ = 100000 - 30240 = 69760 \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

Upvote

Exemplar Solution Class 11th

The number of terms in the expansion of ${[{(2 x + y^{3})}^{4}]}^{7}$ is 8.

4 Views|Posted a year ago

Asked by Vishal Baghel

S

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

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

$\begin{array}{l} G i v e n expression i s {[{(2 x + y^{3})}^{4}]}^{7} = {(2 x + y^{3})}^{28} \\ H e r e, n = 28 \\ ∴ T h e n u m b e r o f t e r m s i n g i v e n expression i s (n + 1) \\ i . e ., 28 + 1 = 29 \\ H e n c e, t h e g i v e n s t a t e m e n t i s' F a l s e' . \end{array}$

Upvote

Exemplar Solution Class 11th

The number of ways in which a team of eleven players can be selected from 22 players always including 2 of them and excluding 4 of them is:

(a) ¹⁶C₁₁

(b) ¹⁶C₅

(c) ¹⁶C₉

(d) ²⁰C₉

8 Views|Posted a year ago

Asked by Payal Gupta

S

Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} T o t a l n u m b e r o f p l a y e r s = 22 \\ 2 p l a y e r s a r e a l w a y s i n c l u d e d a n d 4 a r e a l w a y s e x c l u d i n g o r n e v e r i n c l u d e d = 22 - 2 - 4 = 16 \\ ∴ Required n u m b e r o f s e l e c t i o n = {}_{16}{C_{9}} \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

Upvote

Exemplar Solution Class 11th

The expression $7^{9} + 9^{7}$ is divisible by 64.

[Hint: $7^{9} + 9^{7} = {(1 + 8)}^{9} - {(1 - 8)}^{7}$ .]

8 Views|Posted a year ago

Asked by Vishal Baghel

S

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

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

$\begin{array}{l} 7^{9} + 9^{7} = {(1 + 8)}^{7} - {(1 - 8)}^{9} \\ = [{}_{7}{C_{0}} + {}_{7}{C_{1}} . 8 + {}_{7}{C_{2}} {(8)}^{2} + {}_{7}{C_{3}} {(8)}^{3} + \dots + {}_{7}{C_{7}} {(8)}^{7}] - [\begin{array}{l} {}_{9}{C_{0}} - {}_{9}{C_{1}} . 8 + {}_{9}{C_{2}} {(8)}^{2} - {}_{9}{C_{3}} {(8)}^{3} \\ + \dots {}_{9}{C_{9}} {(8)}^{9} \end{array}] \\ = (7 \times 8 + 9 \times 8) + (21 \times 8^{2} - 36 \times 8^{2}) + \dots \\ = (56 + 72) + (21 - 36) \times 8^{2} + \dots \\ = 128 + 64 (21 - 36) + \dots \\ = 64 [2 + (21 - 36) + \dots] \\ w h i c h i s d i v i s i b l e b y 64 . \\ H e n c e, t h e g i v e n s t a t e m e n t i s' T r u e' . \end{array}$

Upvote

Exemplar Solution Class 11th

The number of parallelograms that can be formed from a set of four parallel lines intersecting another set of three parallel lines is:

(a) 6

(b) 18

(c) 12

(d) 9

4 Views|Posted a year ago

Asked by Payal Gupta

S

Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} W e k n o w t h a t t o f o r m a parallelogram, w e r e q u i r e a p a i r o f l i n e s f r o m a s e t o f 4 l i n e s a n d \\ a n o t h e r p a i r o f l i n e s f r o m a n o t h e r s e t o f 3 l i n e s \\ ∴ T h e r e q u i r e d n u m b e r o f parallelograms = {}_{4}{C_{2}} \times {}_{3}{C_{2}} = 6 \times 3 = 18 \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

Upvote

Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Nine

If $θ_{1}, θ_{2}, θ_{3}, . . ., θ_{n}$ are in A.P., whose common difference is $d$ , show that

2 Views|Posted a year ago

Asked by Alok Kumar Singh

S

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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

Sol:

$\begin{array}{l} Since θ_{1}, θ_{2}, θ_{3}, \dots, θ_{n} a r e i n A . P . \\ ∴ θ_{2} - θ_{1} = θ_{3} - θ_{2} = \dots = θ_{n} - θ_{n - 1} = d \\ N o w w e h a v e t o p r o v e t h a t \\ s e c θ_{1} . s e c θ_{2} + s e c θ_{2} . s e c θ_{3} + \dots + s e c θ_{n - 1} . s e c θ_{n} = \frac{t a n θ_{n} - t a n θ_{1}}{s i n d} \\ L H S . \frac{s i n d}{s i n d} [s e c θ_{1} . s e c θ_{2} + s e c θ_{2} . s e c θ_{3} + \dots + s e c θ_{n - 1} . s e c θ_{n}] \\ T a k i n g o n l y \frac{s i n d [s e c θ_{1} . s e c θ_{2}]}{s i n d} = \frac{s i n d [\frac{1}{c o s θ_{1}} . \frac{1}{c o s θ_{2}}]}{s i n d} \\ = \frac{s i n (θ_{2} - θ_{1})}{s i n d} . \frac{1}{c o s θ_{1} c o s θ_{2}} \\ = \frac{1}{s i n d} [\frac{s i n θ_{2} c o s θ_{1} - c o s θ_{2} s i n θ_{1}}{c o s θ_{1} c o s θ_{2}}] \\ = \frac{1}{s i n d} [\frac{s i n θ_{2} c o s θ_{1}}{c o s θ_{1} c o s θ_{2}} - \frac{c o s θ_{2} s i n θ_{1}}{c o s θ_{1} c o s θ_{2}}] \\ = \frac{1}{s i n d} [t a n θ_{2} - t a n θ_{1}] \\ S i m i l a r l y, w e c a n s o l v e o t h e r t e r m s w h i c h w i l l b e \\ \frac{1}{s i n d} [t a n θ_{3} - t a n θ_{2}] a n d \frac{1}{s i n d} [t a n θ_{4} - t a n θ_{3}] \\ H e r e L H S = \frac{1}{s i n d} [t a n θ_{2} - t a n θ_{1} + t a n θ_{3} - t a n θ_{2} + \dots + t a n θ_{n} - t a n θ_{n - 1}] \\ = \frac{1}{s i n d} [- t a n θ_{1} + t a n θ_{n}] = \frac{t a n θ_{n} - t a n θ_{1}}{s i n d} R H S . \\ L H S = R H S H e n c e p r o v e d . \end{array}$

Upvote

Exemplar Solution Class 11th

The sum of the series ^{10 20} _{r =0} C_ris 2¹⁹ + ²⁰ C₁₀/2

14 Views|Posted a year ago

Asked by Vishal Baghel

S

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

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

$\begin{array}{l} \sum_{r = 0}^{10} {}_{20}{C_{r}} = {}_{20}{C_{0}} + {}_{20}{C_{1}} + {}_{20}{C_{2}} + {}_{20}{C_{3}} + \dots + {}_{20}{C_{10}} \\ = {}_{20}{C_{0}} + {}_{20}{C_{1}} + \dots + {}_{20}{C_{10}} + {}_{20}{C_{11}} + \dots + {}_{20}{C_{20}} - ({}_{20}{C_{11}} + \dots + {}_{20}{C_{20}}) \\ = 2^{20} - ({}_{20}{C_{11}} + \dots + {}_{20}{C_{20}}) \\ H e n c e, t h e g i v e n s t a t e m e n t i s' F a l s e' \end{array}$

Upvote

Exemplar Solution Class 11th

The number of triangles that are formed by choosing the vertices from a set of 12 points, seven of which lie on the same line, is:

(a) 105

(b) 15

(c) 75

(d) 185

2 Views|Posted a year ago

Asked by Payal Gupta

S

Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} T h e t o t a l n u m b e r o f t r i a n g l e s f o r m e d f r o m 12 points t a k i n g 3 a t a t i m e = {}_{12}{C_{3}} . \\ B u t g i v e n t h a t o u t o f 12 points, 7 a r e c o l l i n e a r \\ S o, t h e s e s e v e n points w i l l f o r m n o t r i a n g l e . \\ ∴ T h e r e q u i r e d n u m b e r o f t r i a n g l e s = {}_{12}{C_{3}} - {}_{7}{C_{3}} \\ \Rightarrow \frac{12!}{3! 9!} - \frac{7!}{3! 4!} = \frac{12 \times 11 \times 10 \times 9!}{3 \times 2 \times 1 \times 9!} - \frac{7 \times 6 \times 5 \times 4!}{3 \times 2 \times 1 \times 4!} \\ \Rightarrow \frac{12 \times 11 \times 10}{3 \times 2} - \frac{7 \times 6 \times 5}{3 \times 2} = 220 - 35 = 185 \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

Upvote

Exemplar Solution

Questions

Discussions

Active Users

Followers

If the sum of $p$ terms of an A.P. is $q$ and the sum of $q$ terms is $p$ , show that the sum of $p + q$ terms is $- (p + q)$ . Also, find the sum of the first $p - q$ terms ( $p > q$ ).

The sum of coefficients of the two middle terms in the expansion of ${(1 + x)}^{2 n - 1}$ is equal to ${}_{2 n - 1}{C_{n}}$ .

The number of 5-digit telephone numbers having at least one of their digits repeated is:

(a) 90,000

(b) 10,000

(c) 30,240

(d) 69,760

The number of terms in the expansion of ${[{(2 x + y^{3})}^{4}]}^{7}$ is 8.

The expression $7^{9} + 9^{7}$ is divisible by 64.

[Hint: $7^{9} + 9^{7} = {(1 + 8)}^{9} - {(1 - 8)}^{7}$ .]

The number of parallelograms that can be formed from a set of four parallel lines intersecting another set of three parallel lines is:

(a) 6

(b) 18

(c) 12

(d) 9

If $θ_{1}, θ_{2}, θ_{3}, . . ., θ_{n}$ are in A.P., whose common difference is $d$ , show that

The sum of the series ^{10 20} _{r =0} C_ris 2¹⁹ + ²⁰ C₁₀/2

The number of triangles that are formed by choosing the vertices from a set of 12 points, seven of which lie on the same line, is:

(a) 105

(b) 15

(c) 75

(d) 185