find the sum of the following series up to n terms.5 + 55 + 555 + …..6 + .66 + .666 + ….
hint: first of all, multiply and divide 9 in both the series after taking common 5 and 0.6 from them respectively. then make a g.p of the form \[10,{{10}^{2}},{{10}^{3}}....\] by writing 9 = 10 – 1, 99 = 100 – 1, and so on. finally, use the formula for the sum of n terms of g.p that is \[\dfrac{a\left( 1-{{r}^{n}} \right)}{\left( 1-r \right)}\].complete step-by-step solution -in this question, we have to find the sum of the n terms of the series.(i) 5 + 55 + 555 + ….(ii) .6 + .66 + .666 + ….let us consider the first seriess = 5 + 55 + 555 + ….. n termsby taking out 5 common from the above series, we get, s = 5 (1 + 11 + 111 + 1111 …. n terms)by multiplying 9 on both the sides of the above equation, we get, 9s = 5 [9 + 99 + 999 …… n times]\[\dfrac{9s}{5}=9+99+999.....\text{ n times}\]we can also write the above equation as, \[\dfrac{9s}{5}=\left( 10-1 \right)+\left( 100-1 \right)+\left( 1000-1 \right)....\text{ n times}\]by simplifying the above equation and taking the term 1 separately, we get, \[\dfrac{9s}{5}=\left( {{10}^{1}}+{{10}^{2}}+{{10}^{3}}.....\text{ n times} \right)-\left( 1+1+1......\text{ n times} \right)\]we know that 1 added n times is equal to n and the nth term of 1, 2, 3…. n. so, we get, \[\dfrac{9s}{5}=\left( {{10}^{1}}+{{10}^{2}}+{{10}^{3}}{{....10}^{n}} \right)-\left( n \right)\]we know that the series of the form \[a,ar,a{{r}^{2}},a{{r}^{3}}.....a{{r}^{n}}\] is in g.p and sum of the n terms of this g.p is given by \[\dfrac{a\left( 1-{{r}^{n}} \right)}{\left( 1-r \right)}\]in the above equation, we can see that \[10,{{10}^{2}},{{10}^{3}}....\] are in g.p with the first term and common ratio as 10. so, we get, \[\dfrac{9s}{5}=\dfrac{10\left( 1-{{\left( 10 \right)}^{n}} \right)}{1-10}-n\]\[\dfrac{9s}{5}=\dfrac{10\left( 1-{{\left( 10 \right)}^{n}} \right)}{-9}-n\]\[\dfrac{9s}{5}=\dfrac{10}{9}\left( {{10}^{n}}-1 \right)-n\]\[\dfrac{9s}{5}=\dfrac{10\left( {{10}^{n}}-1 \right)-9n}{9}\]by multiplying \[\dfrac{5}{9}\] on both the sides of the above equation, we get, \[s=\dfrac{5}{9}\left[ \dfrac{10\left( {{10}^{n}}-1 \right)-9n}{9} \right]\]\[s=\dfrac{50\left( {{10}^{n}}-1 \right)}{81}-\dfrac{5}{9}n\]hence, the sum of the sequence 5 + 55 + 555….. up to n terms is equal to \[\dfrac{50\left( {{10}^{n}}-1 \right)}{81}-\dfrac{5n}{9}\]let us consider the second set.s = 0.6 + 0.66 + 0.666 + ….. n times.by taking out 0.6 common, we can write the sequence as, s = 6 [0.1 + 0.11 + 0.111 ….. n terms]by multiplying and dividing rhs by 9, we get,\[s=\dfrac{6}{9}\left[ 0.9+0.99+0.999.....n\text{ terms} \right]\]\[s=\dfrac{2}{3}\left[ \dfrac{9}{10}+\dfrac{99}{100}+\dfrac{999}{1000}+\dfrac{9999}{10000}.....n\text{ terms} \right]\]we can write the above equation as, \[s=\dfrac{2}{3}\left[ \dfrac{\left( 10-1 \right)}{10}+\dfrac{\left( 100-1 \right)}{100}+\dfrac{\left( 1000-1 \right)}{1000}+\dfrac{\left( 10000-1 \right)}{10000}.....n\text{ terms} \right]\]\[s=\dfrac{2}{3}\left[ \left( 1-\dfrac{1}{10} \right)+\left( 1-\dfrac{1}{100} \right)+\left( 1-\dfrac{1}{1000} \right)+\left( 1-\dfrac{1}{10000} \right).....n\text{ terms} \right]\]\[s=\dfrac{2}{3}\left[ \left( 1+1+1......n\text{ terms} \right)-\left( \dfrac{1}{10}+\dfrac{1}{100}+\dfrac{1}{1000}.+...n\text{ terms} \right) \right]\]here, we can see that \[\dfrac{1}{10}+\dfrac{1}{{{\left( 10 \right)}^{2}}}+\dfrac{1}{{{\left( 10 \right)}^{3}}}+.....n\text{ terms}\] is in gp with \[a=\dfrac{1}{10}\] and \[r=\dfrac{1}{10}\].we know that the sum of n terms of the g.p is \[\dfrac{a\left( 1-{{r}^{n}} \right)}{\left( 1-r \right)}\]. so, we get, \[s=\dfrac{2}{3}\left[ \left( n \right)-\dfrac{\dfrac{1}{10}\left( 1-{{\left( \dfrac{1}{10} \right)}^{n}} \right)}{1-\dfrac{1}{10}} \right]\]\[s=\dfrac{2}{3}\left[ n-\dfrac{\dfrac{1}{10}\left( 1-\dfrac{1}{{{10}^{n}}} \right)}{\dfrac{9}{10}} \right]\]\[s=\dfrac{2}{3}\left[ n-\dfrac{1}{9}\left( 1-\dfrac{1}{{{10}^{n}}} \right) \right]\]we know that \[\dfrac{1}{{{a}^{n}}}={{a}^{-n}}\]. by using this, we get, \[s=\dfrac{2}{3}\left( n-\dfrac{1}{9}\left( 1-{{10}^{-n}} \right) \right)\]so, \[s=\dfrac{2}{3}n-\dfrac{2}{27}\left[ 1-{{\left( 10 \right)}^{-n}} \right]\]hence, we get the sum of the sequence 0.6 + 0.66 + 0.666 + ….. n terms as \[\dfrac{2}{3}n-\dfrac{2}{27}\left[ 1-{{\left( 10 \right)}^{-n}} \right]\].note: in this question, students must note that whenever we are in need to find the sum of series of the type a + aa + aaa + aaaa ….., we always have to make the series of 9 + 99 + 999 + ….. type by multiplying and dividing the whole series by 9 so that we can get a g.p of the type \[10,{{10}^{2}},{{10}^{3}}\]…. also in this question, some students make this mistake of considering 5 + 55 + 555 + …. or 0.6 + 0.66 + 0.666 + …. in g.p which is wrong because we can see that there is no common ratio in these sequences.
65 ÷ ( – 13) = **-5** use

summation of series of natural numbers -


 

- wherein

sum of first n natural numbers



 

 and

 

summation of series of natural numbers -


 

- wherein

sum of  squares of first n natural numbers



 

 

3+7+14+24+37 15k

let tn=an2+bn+c            because difference is in a.p: [4,7,10,13]

at n=1     (i)

    n=2      (ii)

    n=3        (iii)

    (ii)-(i)        (iv)

    (iii)-(ii)       (v)

    (v)-(iv)      

                    

 

from (i) 

from (ii) 







 



      

 

put n=15   

               

               

               

               

 




option 1)

126

this option is incorrect

option 2)

122

this option is correct

option 3)

81

this option is incorrect

option 4)

119

this option is incorrect answer: `4+1+(1)/(4)+(1)/(6)+...`
solution: n/a sum of a g.p. series is represented by the formula, , when |r|<1. ‘sn’ represents the sum of the g.p. series upto nth terms, ‘a’ represents the first term, ‘r’ represents the common ratio and ‘n’ represents the number of terms.here,a = 1r = (ratio between the n term and n-1 term) n terms∴ ⇒∴ given that,l ‖ m and n cuts themlet,∠1 = 65o∠2 = x∠3 = 40o∠1 = ∠4 = 65o (alternate angle) (i)∠3 + ∠4 + ∠5 = 180o (angle sum property)40o + 65o + ∠5 = 180o∠5 = 75onow,∠2 + ∠5 = 180o (linear pair)x + 75o = 180ox = 105o (a) for digit



expression : 5n + 1

where n = no of digits

for 5 digits

no of segments required = 5(5) + 1 = 25 + 1 = 26

for 10 digits

no of segments required = 5(10) + 1 = 50 + 1 = 51

for 100 digits

no of segments required = 5(100) + 1 = 500 + 1 = 501

(b) for digit



expression : 3n + 1

where n = no of digits

for 5 digits

no of segments required = 3(5) + 1 = 15 + 1 = 16

for 10 digits

no of segments required = 3(10) + 1 = 30 + 1 = 31

for 100 digits

no of segments required = 3(100) + 1 = 300 + 1 = 301

(c) for digit



expression : 5n + 2

where n = no of digits

for 5 digits

no of segments required = 5(5) + 2 = 25 + 2 = 27

for 10 digits

no of segments required = 5(10) + 2 = 50 + 2 = 52

for 100 digits

no of segments required = 5(100) + 2 = 500 + 2 = 502 let x denote the number of sixes. p(getting a six when a die is thrown) = p = `(1)/(6)` ∴ q = 1 – p = `1 - (1)/(6) = (5)/(6)` given, n = 6∴ x ∼ b`(6, 1/6)`the p.m.f. of x is given by p(x = x) = `""^6"c"\_x (1/6)^x (5/6)^(6 - x), x` = 0, 1,...,6p(getting at most 2 sixes) = p(x ≤ 2) = p(x = 0 or x = 1 or x = 2) = p(x = 0) + p(x = 1) + p(x = 2) = `""^6"c"\_0(1/6)^0 (5/6)^6 + ""^6"c"\_1(1/6)(5/6)5 + ""^6"c"\_2(1/6)^2 (5/4)^4` = `(5/6)^4 + 6(1/6)(5/6)^5 + (6!)/(2! xx 4!)(1/6)^2 (5/6)^4` = `(5/6)^4 [(5/6)^2 + (5/6) + (6 xx 5 xx 4!)/(2 xx 1 xx 4) xx (1/6)^2]` = `(5/6)^4 [25/6^2 + 30/6^ + 15/6^2]` = `(5/6)^4 (70/36)` = `(5/6)^4 (35/18)` = `(5/4)^4 ((5 xx 7)/(6 xx 3))` = `(7)/(3)(5/6)^5`. if the value (or total amount) of these two kinds of coins is thirty dollars..
let n = # of nickles, so 500-n = # of dimes.
 
0.05n + 0.10(500-n) = 30.00 expressed the total value; solve for n and you'll then be able to give each count requested. a sum of money becomes 3 times in 5 years at simple interest. in how many years will the same sum become 6 times at the same rate of simple interest? finding the correct answer here is important, but perhaps equally important is finding it using the method of matrix reduction, since those are the instructions.the augmented matrix is4 -2 -5 -514 -1 -2 -364 -2 4 -24you'll have to forgive the poor formatting. this answer box doesn't have a proper math symbol editor.using elementary row operations, the reduced row echelon form of the augmented matrix is1 0 0 -60 1 0 60 0 1 3from which we conclude that the system has a unique solution (-6,6,3).