Free Study Material

NCERT Solutions

NCERT Books

Revision Notes

Practice Problems

This is one of our many quizzes on Finite arithmetic series. Click on this to access more such competitive quizzes on Algebra.

Finite arithmetic series Quiz 2

mathematics

 The ratio of the sum of the first nine terms of two A.P. series $115:132$ respectively. The first terms of both the series are equal. The ratio of the first term to the common difference of the second series is $20:1$ respectively. The sum of the first thirteen terms of the first series is 611. What is the difference between the common differences?

 By using the nth term formula, ${{a}_{n}}=a+(n-1)d$, we will get an equation containing both the common differences, from there we will equate the difference between them. 

  Let the first term be a and the common differences be d and d' respectively. $ \\ a:d'=20:1,a=20d'  \\   \text{So},\left( a+4d \right):\left( a+4d' \right)=115:132  \\   132a+528d=115a+460d'\Rightarrow 17a+528d=460d'  \\   528d=460d'-340d'=120d'  \\   22d=5d'  \\ \text{And, }13\times \left( a+6d \right)=611  \\   a+6d=47  \\   20d'+\dfrac{30}{22}d'=47\Rightarrow 235d'=47\times 11  \\   d'=\dfrac{11}{5},d=\dfrac{1}{2}  \\   d-d'=\dfrac{11}{5}-\dfrac{1}{2}=\dfrac{17}{10}=1.7 $

 The first term of an A.P. series is 123 while the sum of its first sixteen terms is 1368. What will be the sum of the first six terms of the G.P. series whose first term is 30 and the common ratio is equal to the reciprocal of the common difference of the given A.P. series?

 By using the summation formula, ${{S}_{n}}=\dfrac{n}{2}\left\{ 2a+(n-1)d \right\}$ we will get the common difference, then by the summation formula of the GP series, ${{S}_{n}}=a\left[ \dfrac{({{r}^{n}}-1)}{(r-1)} \right]$ we will get the required sum.

  $a=123,{{S}_{16}}=1368 \\ \Rightarrow 1368=\dfrac{16}{2}(2\times 123+15d) \\ \Rightarrow 15d=171-246=-75,d=-5 \\\text{For the GP series, } \\ a=30,r=-\dfrac{1}{5} \\ {{S}_{6}}=\dfrac{30\left\{ 1-{{\left( -\dfrac{1}{5} \right)}^{6}} \right\}}{\left( 1+\dfrac{1}{5} \right)}=\dfrac{15624}{625}$

 The first terms of three A.P. series are in the ratio $2:4:5$ respectively while their common differences are in the ratio $1:5:7$ respectively. Also, the number of terms in the series are in the ratio $3:5:8$ respectively. What is the sum of terms of the second series if the sum of terms of the first series is 15 and the difference between the first term of the third series and the common difference of the first series is $\dfrac{6}{7}$ and the product of the first term of the first series and the number of terms of the second series is $\dfrac{25}{2}$?

 By using the summation formula, ${{S}_{n}}=\dfrac{n}{2}\left\{ 2a+(n-1)d \right\}$ we will get one equation, and by using the two given expressions we will get two more equations, solving all three will give the number of terms, first term and the common difference, from there we can find the 35th term.

  Let the first terms of the A.P. series be 2a, 4a and 5a respectively and the common differences be $d, 5d $ and $7d$ and the number of terms be $3n, 5n$ and $8n$ respectively. For the first series: $ \\ {{S}_{3n}}=\dfrac{3n}{2}\left\{ 4a+(3n-1)d \right\}\Rightarrow \dfrac{3n}{2}\left\{ 4a+3nd-d \right\}=15 \\ \Rightarrow n(4a+3nd-d)=10...(1) \\ $ Also, $ \\ 2a\times 5n=\dfrac{25}{2}\Rightarrow 4an=5...(2) \\ 5a-d=\dfrac{6}{7}\Rightarrow 35a-7d=6...(3) \\ $ From $(1), (2)$ and $(3)$, we get, $ \\ 5+3{{n}^{2}}\times \dfrac{175-24n}{28n}-\dfrac{175-24n}{28n}n=10\Rightarrow 72{{n}^{2}}-549n+315=0 \\ \Rightarrow n=7,\dfrac{5}{8}\text{,d=}\dfrac{1}{28}\text{ and }a=\dfrac{5}{28} \\ $ For the second series, ${{S}_{35}}=\dfrac{35}{2}\left( 8\times \dfrac{5}{28}+34\times 5\times \dfrac{1}{28} \right)=\dfrac{525}{4}$ 

 The rate of flow of water from three taps are 2, N and 3, $\text{(in liter per hour)}$ where N tap’s efficiency is neither more nor less than other two taps. If the lowest and the highest flow rates are both decreased by a certain quantity x, while the intermedia rate is left unchanged. If the reciprocals of the three flow rates, in the order given above, are in arithmetic progression both before and after the change, then what is the quantity $x \text{ (in liter per hour)}$? $\text{(Negative flow rates indicate that the taps act as emptying taps instead of filling taps)}$

 From the given rates, we can get the efficiencies, then as they form AP, we can equate and get the value for N then from there we can get the value of x by putting the value of n as $\dfrac{1}{\left( 2-x \right)},\dfrac{1}{2.4}\text{ and }\dfrac{1}{\left( \text{3 }-x \right)}$are in AP.

  $\dfrac{1}{2},\dfrac{1}{N}\text{and }\dfrac{1}{3} \text{ are in AP.}  \\ \left( \dfrac{1}{N} \right)-\left( \dfrac{1}{2} \right)=\left( \dfrac{1}{3} \right)-\left( \dfrac{1}{N} \right)  \\   \left( \dfrac{1}{N} \right)+\left( \dfrac{1}{N} \right)=\left( \dfrac{1}{3} \right)+\left( \dfrac{1}{2} \right)  \\   \dfrac{2}{N}=\dfrac{5}{6}  \\ N=\dfrac{12}{5}=2.4\text{ litres per hour} \\ \text{Again, }\dfrac{1}{\left( 2-x \right)},\dfrac{1}{2.4}\text{ and }\dfrac{1}{\left( \text{3 }-x \right)}\text{ are in AP } \\ \dfrac{1}{2.4}-\dfrac{1}{(2-x)}=\dfrac{1}{(3-x)}-\dfrac{1}{2.4}\Rightarrow \dfrac{1}{(3-x)}+\dfrac{1}{(2-x)}=\dfrac{10}{12}  \\   \dfrac{\left( 5-2x \right)}{\left( \left( 3-x \right)\left( 2-x \right) \right)}=\dfrac{10}{12}\Rightarrow 60-24x=60-50x+10{{x}^{2}}  \\   10{{x}^{2}}-26x=0\Rightarrow x=0\text{ or }x=2.6  \\ \text{Since }x\text{ cannot be equal to zero,} \\ \therefore x=2.6\text{liters per hour}$

 There are $\left( 2x+5 \right)$ numbers in arithmetic progression, such that the largest of the numbers is $x$ times the smallest number. If the sum of these numbers is 110 times of $x$, then find the smallest number where x is expressed as the number of terms in an expansion of ${{\left( {{2}^{\frac{1}{3}}}+{{2}^{-\dfrac{1}{2}}} \right)}^{x}}$whose last term is ${{\left[ \dfrac{1}{3\times {{\left( 9 \right)}^{\frac{1}{3}}}} \right]}^{lo{{g}_{3}}8}}$.

 We can equate the last of the expansion given by Binomial theorem for the value of x, then from there by summation formula ${{S}_{n}}=\dfrac{n}{2}\left\{ 2a+(n-1)d \right\}$ we can the required expression.

  Last term of $ {{\left( {{2}^{\frac{1}{3}}}+{{2}^{-\dfrac{1}{2}}} \right)}^{x}}={}^{x}{{C}_{x}}{{\left( {{2}^{\frac{1}{3}}} \right)}^{0}}{{\left( {{2}^{-\dfrac{1}{2}}} \right)}^{x}}={{\left[ \dfrac{1}{3\times {{\left( 9 \right)}^{\frac{1}{3}}}} \right]}^{lo{{g}_{3}}8}} \\ \Rightarrow {{2}^{-\dfrac{x}{2}}}={{\left( {{3}^{-\dfrac{5}{3}}} \right)}^{lo{{g}_{3}}8}}\Rightarrow {{2}^{-\dfrac{x}{2}}}={{\left( {{3}^{-\dfrac{5}{3}}} \right)}^{lo{{g}_{3}}{{2}^{3}}}} \\ \Rightarrow {{2}^{\frac{x}{2}}}={{\left( {{3}^{5}} \right)}^{lo{{g}_{3}}2}}=({{3}^{5{{\log }_{3}}2}}) \\ \Rightarrow {{\log }_{3}}{{2}^{\frac{x}{2}}}=5{{\log }_{3}}2\Rightarrow {{\log }_{3}}{{2}^{\frac{x}{2}}}={{\log }_{3}}{{2}^{5}}\Rightarrow {{2}^{\frac{x}{2}}}={{2}^{5}}\Rightarrow x=10 \\ 2x+5=25\text{ and }110x=1100 \\ $ As there are 25 numbers in AP, $n=25$ , let the smallest number be $m$, so largest number $=nx=10$ <br>Sum of these numbers is $\dfrac{25}{2}\left[ n+10n \right]=\dfrac{25}{2}\times 11n=1100,n=8=x-2$ 