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New answer posted

8 months ago

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

If $1 + (1 - 2^{2} \cdot 1) + (1 - 4^{2} \cdot 3) + (1 - 6^{2} \cdot 5) + \dots \dots . + (1 - 20^{2} \cdot 19) = α - 220 β$ , then an ordered pair $(α, β)$ is equal to:

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alok kumar singh

Contributor-Level 10

$1 + (1 - 2^{2} . 1) + (1 - 4^{2} . 3) + \dots \dots . + (1 - 20^{2} . 19)$

$= α - 220 β$

$= 11 - (2^{2} \cdot 1 + 4^{2} \cdot 3 + \dots . . + 20^{2} \cdot 19)$

$= 11 - 2^{2} \cdot \sum_{r = 1}^{10} ? r^{2} (2 r - 1) = 11 - 4 (\frac{110^{2}}{2} - 35 * 11)$

$= 11 - 220 (103)$

$\Rightarrow α = 11, β = 103$

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New answer posted

8 months ago

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

Let $a_{1}, a_{2}, \dots a_{n}$ be a given A.P. whose common difference is an integer and $S_{n} = a_{1} + a_{2}$ $+ \dots + a_{n}$ . If $a_{1} = 1, a_{1} = 300$ and $15 \leq n \leq 50$ , then the ordered pair $(S_{n - 4}, a_{n - 4})$ is equal to:

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alok kumar singh

Contributor-Level 10

$x \frac{d y}{d x} - y = x^{2} (x c o s ? x + s i n ? x), x > 0$

$\frac{d y}{d x} - \frac{y}{x} = x (x c o s ? x + s i n ? x) \Rightarrow \frac{d y}{d x} + P y = Q$

So, I.F. $= e^{\int - \frac{1}{x} d x} \frac{1}{| x |} = \frac{1}{x} (x > 0)$

Thus, $\frac{y}{x} = \int \frac{1}{x} (x (x c o s ? x + s i n ? x)) d x$

$\Rightarrow \frac{y}{x} = x s i n ? x + C$

$? y (π) = π \Rightarrow C = 1$

So, $y = x^{2} s i n ? x + x \Rightarrow (y)_{π / 2} = \frac{π^{2}}{4} + \frac{π}{2}$

Also, $\frac{d y}{d x} = x^{2} c o s ? x + 2 x s i n ? x + 1$

$\Rightarrow \frac{d^{2} y}{d x^{2}} = - x^{2} s i n ? x + 4 x c o s ? x + 2 s i n ? x$

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New answer posted

8 months ago

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

If a₁, a₂, a₃…. and b₁, b₂, b₃…. are A.P., and a₁ = 2, a₁₀ = 3, a₁b₁ = 1 – a₁₀b₁₀, then a₄b₄ is equal to…….

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Payal Gupta

Contributor-Level 10

$a_{1}, a_{2}, a_{3} . . . . . .$ are in A.P. (Let common difference is d₁)

$b_{1}, b_{2}, b_{3} . . . .$ are in A.P. (Let common difference is d₂)

and a₁2, a₁₀ = 3, a₁b₁ = 1 = a₁₀b₁₀

$? a_{1} b_{1} = 1$

$∴ b_{1} = \frac{1}{2}$

$a_{10} b_{10} = 1$

$∴ b_{10} = \frac{1}{3}$

Now,

$a_{10} = a_{1} + 9 d_{1} \Rightarrow d_{1} = \frac{1}{9}$

$b_{10} = b_{1} + 9 d_{2} \Rightarrow d_{2} = \frac{1}{9} [\frac{1}{3} - \frac{1}{2}] = - \frac{1}{54}$

Now

$a_{4} = 2 + \frac{3}{9} = \frac{7}{3}$

$b_{4} = \frac{1}{2} - \frac{3}{54} = \frac{4}{9}$

$∴ a_{4} b_{4} = \frac{28}{27}$

...more

$a_{1}, a_{2}, a_{3} . . . . . .$ are in A.P. (Let common difference is d₁)

$b_{1}, b_{2}, b_{3} . . . .$ are in A.P. (Let common difference is d₂)

and a₁2, a₁₀ = 3, a₁b₁ = 1 = a₁₀b₁₀

$? a_{1} b_{1} = 1$

$∴ b_{1} = \frac{1}{2}$

$a_{10} b_{10} = 1$

$∴ b_{10} = \frac{1}{3}$

Now,

$a_{10} = a_{1} + 9 d_{1} \Rightarrow d_{1} = \frac{1}{9}$

$b_{10} = b_{1} + 9 d_{2} \Rightarrow d_{2} = \frac{1}{9} [\frac{1}{3} - \frac{1}{2}] = - \frac{1}{54}$

Now

$a_{4} = 2 + \frac{3}{9} = \frac{7}{3}$

$b_{4} = \frac{1}{2} - \frac{3}{54} = \frac{4}{9}$

$∴ a_{4} b_{4} = \frac{28}{27}$

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New answer posted

9 months ago

Sequences and Series Ncert Solutions Maths class 11th

6. Kindly Consider the following a_n = n $\frac{n^{2} + 5}{4}$

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Payal Gupta

Contributor-Level 10

6. Here, a_n=n $\frac{x^{2} + 5}{4}$

Putting n=1,2,3,4,5 we get,

$a_{1} = 1 * \frac{(1^{2} + 5)}{4} = 1 * \frac{6}{4} = \frac{3}{2}$

$a_{2} = 2 * (\frac{2^{2} + 5}{4}) = 2 * \frac{(4 + 5)}{4} = \frac{9}{2}$

$a_{3} = 3 * \frac{(3^{2} + 5)}{4} = 3 * \frac{(9 + 5)}{4} = 3 * \frac{14}{4} = \frac{21}{2}$

$a_{4} = 4 * \frac{(4^{2} + 5)}{4} = 4 * \frac{(16 + 5)}{4} = 4 * \frac{21}{4} = 21$

$a_{5} = 5 * \frac{(5^{2} + 5)}{4} = 5 * \frac{(25 + 5)}{4} = 5 * \frac{30}{4} = \frac{75}{2}$

Hence, the first five terms are $\frac{3}{2}, \frac{9}{2}, \frac{21}{2}, 21, \frac{75}{2}$ .

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New answer posted

9 months ago

Sequences and Series Ncert Solutions Maths class 11th

3. Kindly Consider the following a_n=2ⁿ

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Payal Gupta

Contributor-Level 10

3. Here a_n=2ⁿ

Substituting n=1,2,3,4,5 we get,

a₁=2¹=2

a₂=2²=4

a₃=2³=8

a₄=a⁴=16

a₅=2⁵=32.

Hence the first five terns are 2,4,16,32 and 64.

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New answer posted

9 months ago

Sequences and Series Ncert Solutions Maths class 11th

2. Kindly Consider the following a_n= $\frac{n}{n + 1}$

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Payal Gupta

Contributor-Level 10

2. Here, a₁= $\frac{n}{n + 1}$

Substituting n=1,2,3,4,5 we get,

$a_{1} = \frac{1}{1 + 1} = \frac{1}{2}$

$a_{2} = \frac{2}{2 + 1} = \frac{2}{3}$

$a_{3} = \frac{3}{3 + 1} = \frac{3}{4}$

$a_{4} = \frac{4}{4 + 1} = \frac{4}{5}$

$a_{5} = \frac{5}{5 + 1} = \frac{5}{6}$ .

Hence the first five terns are $\frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \frac{4}{5}$ and $\frac{5}{6}$ .

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New answer posted

10 months ago

Sequences and Series Sequence and Series

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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Payal Gupta

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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New answer posted

10 months ago

Sequences and Series Sequence and Series

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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Payal Gupta

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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New answer posted

10 months ago

Sequences and Series Sequence and Series

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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Payal Gupta

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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New answer posted

10 months ago

Sequences and Series Sequence and Series

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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Payal Gupta

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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