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This is one of our many quizzes on Series basics. Click on this to access more such competitive quizzes on Algebra.

Series basics Quiz 2

mathematics

 Let $\text{a,b,c}$ be three consecutive terms of an G.P. and all greater than $\text{100}$ . Then the value of ${{a}^{\sqrt{{{\log }_{a}}c}}}-{{c}^{\sqrt{{{\log }_{c}}a}}}$ is equal to 

 Considering the term ${{a}^{\sqrt{{{\log }_{a}}c}}}$ and solving it to obtain a relation between ${{a}^{\sqrt{{{\log }_{a}}c}}}$ and ${{c}^{\sqrt{{{\log }_{c}}a}}}$.

  $\text{Consider the term }{{a}^{\sqrt{{{\log }_{a}}c}}} \\ \text{Multiply and divide the exponent with }\sqrt{{{\log }_{a}}c} \\ \Rightarrow \text{ }{{a}^{\sqrt{{{\log }_{a}}c}\times \frac{\sqrt{{{\log }_{a}}c}}{\sqrt{{{\log }_{a}}c}}}}=\,\,\,{{a}^{{{\log }_{a}}c\times \frac{1}{\sqrt{{{\log }_{a}}c}}}} \\ \Rightarrow \,\,\,{{\left( {{a}^{{{\log }_{a}}c}} \right)}^{\frac{1}{\sqrt{{{\log }_{a}}c}}}} \\ \Rightarrow \,\,\,\,{{\left( c \right)}^{\sqrt{\frac{1}{{{\log }_{a}}c}}}} \\ \Rightarrow {{c}^{\sqrt{{{\log }_{c}}a}}} \\ \therefore {{a}^{\sqrt{{{\log }_{a}}c}}}-{{c}^{\sqrt{{{\log }_{c}}a}}}\text{ =0 }$

 If $a=\sum\limits_{i=1}^{6}{\dfrac{1}{\left( 2i+1 \right)\left( 2i+3 \right)}}$. Then find the value of $1+a+{{a}^{2}}+{{a}^{3}}+....\infty $

 Try solving the term $\sum\limits_{i=1}^{6}{\dfrac{1}{\left( 2i+1 \right)\left( 2i+3 \right)}}$ and find the value of $a$  by using the concept of Telescopic series and then use formula to calculate sum of infinite terms in a G.P.

  $\sum\limits_{n=1}^{6}{\dfrac{1}{\left( 2n+1 \right)\left( 2n+3 \right)}}=\sum\limits_{n=1}^{6}{\left( \dfrac{\dfrac{1}{2}}{2n+1}-\dfrac{\dfrac{1}{2}}{2n+3} \right)} \\ \text{It is a Telescopic series. Hence,we get} \\ a=\sum\limits_{n=1}^{6}{\left( \dfrac{\dfrac{1}{2}}{2n+1}-\dfrac{\dfrac{1}{2}}{2n+3} \right)} \\ \Rightarrow \dfrac{\dfrac{1}{2}}{2\left( 1 \right)+1}-\dfrac{\dfrac{1}{2}}{2\left( 6 \right)+3} \\ \therefore a=\dfrac{2}{15} \\ \text{Now, }1+a+{{a}^{2}}+{{a}^{3}}+....\infty \text{ is the sum of infinite terms in G.P. i.e., }\dfrac{a}{1-r} \\ \Rightarrow \dfrac{1}{1-\dfrac{2}{15}} =\dfrac{15}{13}$

 If one of the roots of equation ${{x}^{2}}+2\left( a-3 \right)x+9=0$ is $-3$ and $2,{{a}_{1}},{{a}_{2}},...,{{a}_{20}},a$ are in A.P. Then find the value of ${{a}_{3}}$

 Substituting $-3$ in the given equation, find the value of $a$ and the common difference of the A.P. Then find the value of ${{a}_{3}}$.

  $\text{Given }-3\text{ is a root } \\ {{\left( -3 \right)}^{2}}+2\left( a-3 \right)\left( -3 \right)+9=0 \\ \Rightarrow 9-6\left( a-3 \right)+9=0 \\ \Rightarrow a-3=3 \Rightarrow a=6 \\ \Rightarrow  2,{{a}_{1}},{{a}_{2}},...,{{a}_{20}},6\,\,\text{are in A.P. }\\ \text{Common difference, } \\ d=\dfrac{6-2}{21}=\dfrac{4}{21} \\ {{a}_{3}}=a+3d=2+3\left( \dfrac{4}{21} \right) \\ {{a}_{3}}=\dfrac{18}{7}$

 Let $f\left( x \right)\text{=1+}\dfrac{x}{2}\text{+}\dfrac{{{x}^{2}}}{4}\text{+}\dfrac{{{x}^{3}}}{8}\text{+}............\text{,}$ for $-1\le x\le 1$. Find $\sqrt{{{e}^{\int_{0}^{1}{f\left( x \right)dx}}}}$

 Using a sum of infinite terms in G.P, find the value of $f\left( x \right)$ and then integrate it to obtain the answer.

  $f\left( x \right)=\text{1+}\dfrac{x}{2}\text{+}\dfrac{{{x}^{2}}}{4}\text{+}\dfrac{{{x}^{3}}}{8}\text{+}............ \\ \text{Observe that }f\left( x \right)\text{ is merely an infinite Geometric series with common ratio }r=\dfrac{x}{2} \\ \text{We know that sum of infinite terms in G.P. is }\dfrac{a}{1-r} \\ \Rightarrow f\left( x \right)=\dfrac{a}{1-r}\,=\dfrac{1}{1-\dfrac{x}{2}} \\ \Rightarrow \,f\left( x \right)=\dfrac{2}{2-x} \\ \text{Then }\sqrt{{{e}^{\int_{0}^{1}{f\left( x \right)dx}}}}=\sqrt{{{e}^{\int_{0}^{1}{\left( \dfrac{2}{2-x} \right)dx}}}} \\ \text{Now ,  }\int_{0}^{1}{\left( \dfrac{2}{2-x} \right)}dx=-2\left( \ln \left( 2-x \right) \right)_{0}^{1}=\,-2\left( \ln \left( 1 \right)-\ln \left( 2 \right) \right)=2\ln 2 \\ \therefore \text{ }\sqrt{{{e}^{\int_{0}^{1}{f\left( x \right)dx}}}}=\sqrt{{{e}^{2\ln 2}}}=\sqrt{{{e}^{\ln 4}}}=\sqrt{4}=2$

 The terms of the series${{a}_{1}},{{a}_{2}},...,{{a}_{n}}\,$are determined recursively as follows ${{a}_{1}}=2,$${{a}_{n}}=3{{a}_{n-1}}+1,n\ge 2.$ . If then ${{a}_{n}}-{{a}_{n-1}}=p.{{q}^{n-2}}$ find $p-q$

 Try adding ${{a}_{2}},...,{{a}_{n}}\,$ and subtracting using ${{a}_{n}}=3{{a}_{n-1}}+1$ to find a relation between ${{a}_{n}}$, ${{a}_{n-1}}$. Then by comparing, find the values of $p$ and $q$.

  $\text{We have }{{a}_{n}}=3{{a}_{n-1}}+1 \\ \text{Putting }n=2,3,.....,\text{ successively} \\ {{a}_{2}}=3{{a}_{1}}+1 \\ {{a}_{3}}\,=\,3{{a}_{2}}+1, ..................... {{a}_{n}}=3{{a}_{n-1}}+1 \\ \text{Adding them we get} \\ {{\text{a}}_{2}}+{{a}_{3}}+...+{{a}_{n}}=3\left( {{a}_{1}}+{{a}_{2}}+...+{{a}_{n-1}} \right)+\left( n-1 \right) \\ \Rightarrow S-{{a}_{1}}=3\left( S-{{a}_{n}} \right)+\left( n-1 \right) \\ \Rightarrow \,S=\dfrac{1}{2}\left( 3{{a}_{n}}-{{a}_{1}}-\left( n-1 \right) \right) \\ {{a}_{n}}=3{{a}_{n-1}}+1 \\ {{a}_{n-1}}=3{{a}_{n-2}}+1 \\ \text{Subtracting }{{\text{a}}_{n}}-{{a}_{n-1}}=3\left( {{a}_{n-1}}-{{a}_{n-2}} \right) \\ \text{Let }{{b}_{n}}=3{{b}_{n-1}}\,,n\ge 3 \\ \text{The series }\left\{ {{b}_{n}} \right\}\text{ is a G.P., so } \\ {{b}_{n}}={{3}^{n-2}}{{b}_{2}} \\ \Rightarrow {{b}_{n}}={{3}^{n-2}}\left( {{a}_{2}}-{{a}_{1}} \right) \\ \therefore \text{ }{{a}_{n}}-{{a}_{n-1}}={{5.3}^{n-2}} \\ p=5;\,q=3\,\,\Rightarrow p-q=2$