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Ncert Solutions Maths class 11th

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Ncert Solutions Maths class 11th Sequence and Series

47. Given a G.P. with a = 729 and 7th term 64, determine S₇.

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Payal Gupta

Contributor-Level 10

47. Given, a= 729

$a_{7} = a r^{6} = 64$

$\Rightarrow$ 729. r⁶ = 64

$\Rightarrow$ $r^{6} = \frac{64}{729}$

$\Rightarrow$ $r^{6} = {(\frac{2}{3})}^{6}$

$\Rightarrow$ $r = + \frac{2}{3}$ <1

When $r = \frac{2}{3}$

= $s_{7} = \frac{a (1 - r^{7})}{4 - r} = \frac{729 (1 - {(\frac{2}{3})}^{7})}{1 - \frac{2}{3}} = \frac{729 [1 - \frac{128}{2187}]}{\frac{3 - 2}{2}}$

= $729 * \frac{3}{1} * [\frac{2187 - 128}{2187}]$

= $729 * \frac{3 * 2059}{2187}$

= 2059

When $r = \frac{2}{3}$

$s_{7} = \frac{a (1 - r^{7})}{1 - r} = \frac{729 [1 - {(\frac{- 2}{3})}^{7}]}{1 - (\frac{- 2}{3})} = \frac{729 [1 + \frac{128}{2187}]}{1 + \frac{2}{3}}$

= $\frac{729 [\frac{2187 + 128}{2187}]}{\frac{3 + 2}{3}}$

= $729 * \frac{3}{5} * [\frac{2315}{2187}]$

= 463

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Ncert Solutions Maths class 11th Sequence and Series

46. The sum of first three terms of a G.P. is 16 and the sum of the next three terms is 128. Determine the first term, the common ratio and the sum to n terms of the G.P.

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Payal Gupta

Contributor-Level 10

46. Let a be the first term and r be the common ratio of the G.P.

Then $a_{1} + a_{2} + a_{3} = 16$

$a + a r + a r^{2} + 1 6^{}$

$a (1 + r + r^{2}) = 16$ ………. I

Also $a_{4} + a_{5} + a_{6} = 128$

$a r^{3} + a r^{4} + a r^{5} = 128$

$a r^{3} (1 + r + r^{2}) = 128$ …… II

Dividing II and I we get,

$\frac{a r^{3} (1 + r + r^{2})}{a (1 + r + r^{2})} = \frac{128}{16}$

r³= 8

r³ = 2³

r = 2 >1

So, putting r = 2 in I we get,

$a (1 + 2 + 2^{2}) = 16$

$\Rightarrow$ $a (1 + 2 + 4) = 16$

$\Rightarrow$ $a (7) = 16$

$\Rightarrow$ $a = \frac{16}{7}$

And $s_{x} = \frac{a (r^{n} - 1)}{r - 1}$

$\Rightarrow$ $\frac{\frac{16}{7} (2^{n} - 1)}{2 - 1}$

$\Rightarrow$ $\frac{16}{7} (2^{n} - 1)$

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Ncert Solutions Maths class 11th Sequence and Series

45. How many terms of G.P. 3, 3², 3³, … are needed to give the sum 120?

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Payal Gupta

Contributor-Level 10

45. Given, a=3

$r = \frac{3^{3}}{3} = 3$ >1

If s_n=120

Then $\frac{a (r^{n} - 1)}{r - 1} = 120$

$\Rightarrow$ $\frac{3 (3^{n} - 1)}{3 - 1} = 120$

$\Rightarrow$ $\frac{3 (3^{n} - 1)}{2} = 120$

$\Rightarrow$ $3^{n} - 1 = 120 * \frac{2}{3}$

$\Rightarrow$ 3ⁿ= 80+1

$\Rightarrow$ 3ⁿ= 81

= 3ⁿ = 34

$∴ r = 4$

So, 120 is the sum of first 4 terms of the G.P

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Ncert Solutions Maths class 11th Sequence and Series

44. The sum of first three terms of a G.P. is $\frac{39}{10}$ and their product is 1. Find the common ratio and the terms.

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Payal Gupta

Contributor-Level 10

44. Let the three terms of the G.P be $\frac{a}{r}, a, a r .$

So, $\frac{a}{r} a . a r = 1$

= $a^{3} = 1$

= $a = 1$

And $\frac{a}{r} + a + a r = \frac{39}{10}$

= $\frac{1}{r} + 1 + r = \frac{39}{10}$ (as a = 1)

= $\frac{1 + r + r^{2}}{r} = \frac{39}{10}$

= $(r^{2} + r + 1) * 10 = 39 * r$

= $10 r^{2} + 10 r + 10 = 39 r$

= $10 r^{2} + 10 r - 39 r + 10 = 0$

= $10 r^{2} - 29 r + 10 = 0$

= $10 r^{2} - 25 r - 4 r + 10 = 0$

= $5 r (2 r - 5) - 2 (2 r - 5) = 10$

$= (2 x - 5) (5 r - 2) = 0$

= $(2 x - 5) = 0$ or $(5 r - 2) = 0$

= $r = \frac{5}{2}$ or $r = \frac{2}{5}$

When $a = 1,$ $r = \frac{5}{2}$ the terms are,

$\frac{1}{\frac{5}{2}}, 1,$ $1 * \frac{5}{2} = \frac{2}{5}, 1, \frac{5}{2}$

When $a = 1,$ $r = \frac{2}{5}$ the terms are,

$\frac{1}{\frac{2}{5}},$ 1, $1 * \frac{2}{5} = \frac{5}{2},$ 1, $\frac{2}{5}$

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Ncert Solutions Maths class 11th Sets

43. Evaluate $\sum_{k = 1}^{11} (2 + 3^{x}) .$

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Payal Gupta

Contributor-Level 10

43. $\sum_{}^{11} (2 + 3^{x}) = (2 + 3^{1}) + (2 + 3^{2}) + . . . . . . . . . + (2 + 3^{11})$

$x = 1$ [2+2+ ….+ (11lines)] + (31+ 32+……+311)

= $11 * 2 + \frac{3 (3^{11} - 1)}{3 - 1} [? s_{n} = \frac{a (r^{n} - 1)}{r - 1}, r \geq 1]$

= $22 + \frac{3 (3^{11} - 1)}{2}$

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Ncert Solutions Maths class 11th Sequence and Series

41. 1, – a, a², – a³, . n terms (if a ¹ – 1).

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Payal Gupta

Contributor-Level 10

41. Here, $a_{1} = 1$

$r = \frac{- 9}{1} = - a$

So, $s_{n} = \frac{a_{1} [1 - r^{n}]}{1 - r}$

= $\frac{1 [1 - {(- a)}^{n}]}{1 - (- a)}$

= ${\frac{1 - (a)}{1 + a}}^{n}$

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Ncert Solutions Maths class 11th Sequence and Series

40. √7 , √21 , 3√7 , . n terms.

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Ncert Solutions Maths class 11th Sequence and Series

Find the sum to indicated number of terms in each of the geometric progressions inExercises 39 to 42:

39. 0.15, 0.015, 0.0015, . 20 terms.

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Payal Gupta

Contributor-Level 10

39. Here a = 0.15 = $\frac{15}{100}$

$r = \frac{0.015}{0.15} = \frac{\frac{15}{1000}}{\frac{15}{150}} = \frac{1}{10}$ <1

n = 20

So, Sum to term of G.P., $s_{n} = \frac{a (1 - r^{n})}{1 - r}$

$\Rightarrow$ $s_{20} = \frac{\frac{15}{100} [1 - {(\frac{1}{10})}^{20}]}{1 - \frac{1}{10}}$

= $\frac{\frac{15}{100} [1 - {(0.1)}^{20}]}{\frac{9}{10}}$

= $\frac{15}{100} * \frac{10}{9} [1 - {(0.1)}^{20}]$

= $\frac{1}{6} [1 - {(0.1)}^{20}]$

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Ncert Solutions Maths class 11th Sets

64. In a survey it was found that 21 people liked product A, 26 liked product B and29 liked product C. If 14 people liked products A and B, 12 people liked productsC and A, 14 people liked products B and C and 8 liked all the three products.Find how many liked product C only.

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Payal Gupta

Contributor-Level 10

64. Let A, B and C be the set of people who like product A, B and C respectively.

Then,

Number of people who like product A, n (A) = 21

Number of people who like product B, n (B) = 26

Number of people who like product C, n (C) = 29.

Number of people likes both product A and B, n (A $\cap$ B) = 14

Number of people likes both product A and C, n (A $\cap$ C) = 12

Number of people likes both product B and C, n (B $\cap$ C) = 14.

No. of people who likes all product, n (A $\cap$ B $\cap$ C) = 8

a→n (A $\cap$ B)

b→n (A $\cap$ C)

d→n (B $\cap$ C)

c→n (A $\cap$ B $\cap$ C)

From the above venn diagram we can see that,

Number of people who likes product C only

= n (C) - b - d + c

= n (C) - n (A $\cap$ C) - n (B $\cap$ C) + n (A $\cap$ B $\cap$ C)

= 29 - 12 - 14 +

...more

64. Let A, B and C be the set of people who like product A, B and C respectively.

Then,

Number of people who like product A, n (A) = 21

Number of people who like product B, n (B) = 26

Number of people who like product C, n (C) = 29.

Number of people likes both product A and B, n (A $\cap$ B) = 14

Number of people likes both product A and C, n (A $\cap$ C) = 12

Number of people likes both product B and C, n (B $\cap$ C) = 14.

No. of people who likes all product, n (A $\cap$ B $\cap$ C) = 8

a→n (A $\cap$ B)

b→n (A $\cap$ C)

d→n (B $\cap$ C)

c→n (A $\cap$ B $\cap$ C)

From the above venn diagram we can see that,

Number of people who likes product C only

= n (C) - b - d + c

= n (C) - n (A $\cap$ C) - n (B $\cap$ C) + n (A $\cap$ B $\cap$ C)

= 29 - 12 - 14 + 8

= 11

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Ncert Solutions Maths class 11th Sets

63. In a survey of 60 people, it was found that 25 people read newspaper H, 26 readnewspaper T, 26 read newspaper I, 9 read both H and I, 11 read both H and T,8 read both T and I, 3 read all three newspapers. Find:

(i) The number of people who read at least one of the newspapers.

(ii) The number of people who read exactly one newspaper

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Payal Gupta

Contributor-Level 10

63. Let H, T and I be of people who reads newspaper H, T and I respectively.

Then,

number of people who reads newspaper H, n (H) = 25.

number of people who T, n (T) = 26.

number of people who I, n (I) = 26

number of people who both H and T, n (H $\cap$ I) = 9

number of people who both H and T, n (H $\cap$ T) = 11

number of people who both T and I, n (T $\cap$ I) = 8

number of people who reads all newspaper, n (H $\cap$ T $\cap$ I) = 3.

Total no. of people surveyed = 60

The given sets can be represented by venn diagram

(i) The number of people who reads at least one of the newspaper.

in (H∪T $\cup$ I) = n (H) + n (T) + n (I) n (H $\cap$ T) n (H $\cap$ I) n (T $\cap$ I) + n (H $\cap$ T $\cap$ I)

= 25 + 26 + 26 11 9 8 + 3

= 80 2

...more

63. Let H, T and I be of people who reads newspaper H, T and I respectively.

Then,

number of people who reads newspaper H, n (H) = 25.

number of people who T, n (T) = 26.

number of people who I, n (I) = 26

number of people who both H and T, n (H $\cap$ I) = 9

number of people who both H and T, n (H $\cap$ T) = 11

number of people who both T and I, n (T $\cap$ I) = 8

number of people who reads all newspaper, n (H $\cap$ T $\cap$ I) = 3.

Total no. of people surveyed = 60

The given sets can be represented by venn diagram

(i) The number of people who reads at least one of the newspaper.

in (H∪T $\cup$ I) = n (H) + n (T) + n (I) n (H $\cap$ T) n (H $\cap$ I) n (T $\cap$ I) + n (H $\cap$ T $\cap$ I)

= 25 + 26 + 26 11 9 8 + 3

= 80 28

= 52

(ii) From venn diagram,

a = number of people who reads newspapers H and T only.

b = number of people who reads newspapers H and I only.

c = number of people who reads newspapers T and I only.

d = number of people who reads all newspaper.

So, n (H $\cap$ T) = a + d.

n (H $\cap$ I) = b + d

n (T $\cap$ I) = c + d

So, a + d + c + d + b + d = n (H $\cap$ T) + n (H $\cap$ I) + n (T $\cap$ I)

a + d + c + b + 2d = 11 + 9 + 8

a + b + c +d = 28 - 2d

= 28 2 3 [d = n (H $\cap$ T $\cap$ I) = 3]

= 28 - 6

= 22

Number of people who reads exactly one newspaper

= Total no. of people - No. of people who reads more than one newspaper

= 52 - (a + b +c + d)

= 52 - 22

= 30

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