find the missing term in the given sequence:15, 16, 20, 29, 45, 70, ?&#8230;.
find the missing term in the given sequence:15, 16, 20, 29, 45, 70, ?
find the missing term in the given sequence:15, 16, 20, 29, 45, 70, ?
31, 87, 32, 21 class intervals
frequency
(fi)
xi
di = xi - a
fi di


0 - 20
17
10
-40
-680


20 - 40
28
30
-20
-560


40 - 60
32
50 = a
0
0


60 - 80
p
70
20
20p


80 - 100
19
90
40
760


total
Î£
fi = 96 + p


Î£ fi di = 20p -
480


![](https://s3mn%20mnimgs%20com/img/shared/discuss_editlive/5130830/2013_07_08_13_01_36/mathmlequation8813644107607516415%20png) to find: limitsnote: first check the form of imit. used this method if the limit is satisfying any one from 7 indeterminate form.in this case, indeterminate form is formula used: = 1so, by using the above formula, we have = = therefore, = information is below. you are missing a semicolon at the end :) 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. 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). find and correct the errors in the following mathematical statements

q19) grammer lesson! "their" not there. find an internet phone book. 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