Sequence and Series

Sequence and Series Maths NCERT Exemplar Solutions Class 12th Chapter Nine

Contributor-Level 10

$\frac{6}{3^{12}} + 10 (\frac{1}{3^{11}} + \frac{2}{3^{10}} + \frac{2^{2}}{3^{9}} + \frac{2^{3}}{3^{8}} + . . . . + \frac{2^{10}}{3})$

$\frac{6}{3^{12}} = \frac{10}{3^{11}} (\frac{6^{11} - 1}{6 - 1})$

$= 2^{12} . 1 m . n = 12$

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If is equal to 2ⁿ. m, where m is odd, then n + m is equal to………

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Sequence and Series Maths NCERT Exemplar Solutions Class 12th Chapter Nine

Contributor-Level 10

$1 + (1 + 2^{49}) (2^{49} - 1) = 2^{98}$

M = 1, n = 98

M + n = 99

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Le aa₁, a₂, a₃,…… be an A.P. If $\sum_{r = 1}^{\infty} \frac{a_{r}}{2^{r}} = 4,$ then 4a₂ is equal to……

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Sequence and Series Maths NCERT Exemplar Solutions Class 12th Chapter Nine

Contributor-Level 10

$\sum_{r = 1}^{\infty} \frac{a + (r - 1) d}{2^{r}} = 4, 4 (a + d) = ?$

$\Rightarrow a \sum_{r = 1}^{\infty} {(\frac{1}{2})}^{r} + d \sum_{r = 1}^{\infty} (r - 1) {(\frac{1}{2})}^{r} = 4$

$= {(\frac{1}{2})}^{2} \sum_{k = 0}^{\infty} k {(\frac{1}{2})}^{k - 1}$

$= \frac{1}{4} . \frac{1}{{(1 - \frac{1}{2})}^{2}}$

= 1

a + d = 4

4 (a + d) = 16

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Let S = {1, 2, 3,…., 2022}. Then the probability, that a randomly chosen number n from the set S such that HCF (n, 2022) = 1, is:

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Sequence and Series Maths NCERT Exemplar Solutions Class 12th Chapter Nine

Contributor-Level 10

S = {1,2, 3, …., n, 2022}

HCF (n, 2022) = 1

2022 = 2 * 1011 ->3 * 337

2022 = 2 * 3 * 337 (prime factorization)

Let n (A) = no members divisible by 2 = 1011

Let n (B) = no members divisible by 3 = 674

Let n (C) = no members divisible by 337 = 6

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

= 1011 + 674 + 6 – 337 – 2 – 3 + 1

= 1350

$n (A \cap B) = 337$

$n ((A \cup B \cup C)') = 2022 - 1350 = 672$

Prob. $= \frac{672}{2022} = \frac{336}{1011} = \frac{112}{337}$

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If $\underset{x \to 0}{l i m} \frac{α e^{x} + β e^{- x} + γ s i n x}{x s i n^{2} x} = \frac{2}{3},$ where $α, β, γ \in R,$ then which of the following is NOT correct?

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Contributor-Level 10

$\underset{x \to 0}{l i m} \frac{α e^{x} + β e^{- x} + γ s i n x}{x s i n^{2} x}$

$= \underset{x \to 0}{l i m} \frac{α e^{x} + β e^{- x} + γ s i n x}{x^{3}}$

$\Rightarrow α + β = 0, α - β + γ = 0, \frac{α + β}{2} = 0, \frac{α}{6} - \frac{β}{6} - \frac{γ}{6} = \frac{2}{3}$

$\Rightarrow β = - α γ = - 2 α α - β - γ = 4 \Rightarrow α + α + 2 α = 4 \Rightarrow γ = 1$

$α = 1, β = - 1, γ = - 2$

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4 months ago

106. 150 workers were engaged to finish a job in a certain number of days. 4 workers dropped out on second day, 4 more workers dropped out on third day and so on. It took 8 more days to finish the work. Find the number of days in which the work was completed.

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Contributor-Level 10

106. Let 'x' be the no of days in which 150 workers took to finish the job.

If 150 workers worked for x days then number of workers for x days =150 x.

But given that number of works dropped 4 on 2^nd day, then 4 on 3^rd day and so on taking 8 more

days to finish the work. i.e., x + 8 days we can express as.

150 x = 150 + (150 $-$ 4) + (150 $-$ 4 $-$ 4)+……+ (x + 8) days.

150 x = 150 + 146 + 142 +……… (x+8) days which

R.H.S. from as A.P. of

a = 150

d = -4 and n = x +8

So, S_n = 150 x

$\Rightarrow \frac{n}{2} [2 * 150 + (n - 1) (- 4)] = 150 x$

n [ 150 + (n - 1) (-2)] = 150 (n - 8) [ $?$ n = x +8 x $-$ 8 $-$ x]

150n 2n (n - 1) 150n 1200

2n² + 2n 1200 =0

n² - n - 6

...more

106. Let 'x' be the no of days in which 150 workers took to finish the job.

If 150 workers worked for x days then number of workers for x days =150 x.

But given that number of works dropped 4 on 2^nd day, then 4 on 3^rd day and so on taking 8 more

days to finish the work. i.e., x + 8 days we can express as.

150 x = 150 + (150 $-$ 4) + (150 $-$ 4 $-$ 4)+……+ (x + 8) days.

150 x = 150 + 146 + 142 +……… (x+8) days which

R.H.S. from as A.P. of

a = 150

d = -4 and n = x +8

So, S_n = 150 x

$\Rightarrow \frac{n}{2} [2 * 150 + (n - 1) (- 4)] = 150 x$

n [ 150 + (n - 1) (-2)] = 150 (n - 8) [ $?$ n = x +8 x $-$ 8 $-$ x]

150n 2n (n - 1) 150n 1200

2n² + 2n 1200 =0

n² - n - 600 = 0 [ dividing by -2 throughout]

n² - 25 n + 24 n 600 = 0

n (n - 25) + 24 (n - 25) = 0

(n - 25) (n + 24) = 0

n = 25 or n = -24 (which is not possible)

So, n = 25 days

i.e. x + 8 = 25 days

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105. A manufacturer reckons that the value of a machine, which costs him Rs. 15625, will depreciate each year by 20%. Find the estimated value at the end of 5 years.

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Contributor-Level 10

105. Given,

Cost of machine =? 15625

depreciation rate = 20 % each year.

We have,

Depreciated value after 1st year =? 15625 - 20 % of 15625

=? $15625 - \frac{20}{100} *$ ?15625

=? $15625 (1 - \frac{20}{100})$

=? $15625 * (1 - \frac{1}{5})$

=? $15625 * \frac{4}{5}$

Similarly,

Depreciated value after 2nd year =? 15625 $* {(\frac{4}{5})}^{2}$ and so on.

This, Depreciated value at end of 5 years

=? 15625 $* {(\frac{4}{5})}^{5}$

=? 15625 $* \frac{1024}{3125}$

=? 5120

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104. A man deposited Rs 10000 in a bank at the rate of 5% simple interest annually. Find the amount in 15th year since he deposited the amount and also calculate the total amount after 20 years.

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Contributor-Level 10

104. Given,

Principal amount =? 10000

Amount at end of 1 year =? $(10000 + \frac{10000 + 5 * 1}{100})$

=? (10000 + 500)

=? 10500 {Amount paid = principal + S.I. in a year}

$S . I = \frac{Principal*rate *time}{100}$

Amount at end of 2nd year

=? $(10000 + \frac{10000 * 5 * 2}{100}) {\frac{Principal* rate*time}{100}}$

=? (10000 + 1000)10500 + 500

=? 11000

Amount at end of 3rd year

=? $(10000 + \frac{10000 * 5 * 3}{100})$

=? (1000 + 1500)

=? 11500

So, amount at end of 1^st, 2^nd, 3^rd ………, n^th year forms as A.P.

i.e.? 10500? 11000? 115000, ………. with

a = 10500

d = 11000 - 10500 = 500

Now, Amount in 15th year = Amount at end of 14th year

=? 10500 + ( 14 - 1) * 500

=? 10500 + 13 * 500

=? 10500 + 6500

=? 17000

Similarly amount after 20th year, =? 10500+ (20 - 1) * 500

=? 10500 + 19 * 500

=? 10500 + 9500

...more

104. Given,

Principal amount =? 10000

Amount at end of 1 year =? $(10000 + \frac{10000 + 5 * 1}{100})$

=? (10000 + 500)

=? 10500 {Amount paid = principal + S.I. in a year}

$S . I = \frac{Principal*rate *time}{100}$

Amount at end of 2nd year

=? $(10000 + \frac{10000 * 5 * 2}{100}) {\frac{Principal* rate*time}{100}}$

=? (10000 + 1000)10500 + 500

=? 11000

Amount at end of 3rd year

=? $(10000 + \frac{10000 * 5 * 3}{100})$

=? (1000 + 1500)

=? 11500

So, amount at end of 1^st, 2^nd, 3^rd ………, n^th year forms as A.P.

i.e.? 10500? 11000? 115000, ………. with

a = 10500

d = 11000 - 10500 = 500

Now, Amount in 15th year = Amount at end of 14th year

=? 10500 + ( 14 - 1) * 500

=? 10500 + 13 * 500

=? 10500 + 6500

=? 17000

Similarly amount after 20th year, =? 10500+ (20 - 1) * 500

=? 10500 + 19 * 500

=? 10500 + 9500

=? 20,000

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103. A person writes a letter to four of his friends. He asks each one of them to copy the letter and mail to four different persons with instruction that they move the chain similarly. Assuming that the chain is not broken and that it costs 50 paise to mail one letter. Find the amount spent on the postage when 8th set of letter is mailed.

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Contributor-Level 10

103. As the number of letters mailed forms a G.P. i.e., 4, 4², ……., 4⁸

We have,

a = 4

$r = \frac{4^{2}}{4} = 4$

and n = 8

So, Total numbers of letters = 4 + 4³ + ……. + 4⁸

$= \frac{4 (4^{8} - 1)}{4 - 1}$

$= \frac{4 * (65536 - 1)}{3}$

$= \frac{4}{3} * 65535$

$= \frac{4}{8} * 21, 845$

= 87380

As amount spent on one postage = 50 paise $=$ ? $\frac{50}{100}$

So, for reqd. postage =? $\frac{50}{100} * 87380$

=? 43690

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4 months ago

102. Shamshad Ali buys a scooter for Rs 22000. He pays Rs 4000 cash and agrees to pay the balance in annual instalment of Rs 1000 plus 10% interest on the unpaid amount. How much will the scooter cost him?

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