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Finite geometric series Quiz 2

mathematics

 Fifth term of a G.P. is $\dfrac{4}{{27}}$ and its third term is $\dfrac{4}{3}$. Write its first term.

 Using the formula for the nth term of a G.P; get the value of $r$ and then substitute it in the equation to get the first term.

Given, $ {a_5} = a{r^4} = \dfrac{4}{{27}}\,....(i)$ $ \\ {a_3} = a{r^2} = \dfrac{4}{3}\,....(ii) $  $ \\ {\text{Dividing equation }}\left( {\text{i}} \right){\text{ by }}\left( {{\text{ii}}} \right) \\   \Rightarrow \dfrac{{a{r^4}}}{{a{r^2}}} = \dfrac{{\left( {\dfrac{4}{{27}}} \right)}}{{\left( {\dfrac{4}{3}} \right)}} \Rightarrow {r^2} = \dfrac{1}{9} \Rightarrow r =  \pm \dfrac{1}{3} \\  {\text{Substitute value of }}r{\text{ in equation }}\left( {{\text{ii}}} \right) \\   \Rightarrow a{\left( { \pm \dfrac{1}{3}} \right)^2} = \dfrac{4}{3} \\   \Rightarrow \dfrac{a}{9} = \dfrac{4}{3} \\   \Rightarrow a = 12 \\  \text{Hence, First term of AP is 12}$ 

 write the sum of the G.P. $\sqrt {17} ,\sqrt {68,} \sqrt {272} ,\sqrt {1088} ....$ up to$n$ terms.

 Find the value of common ratio $(r)$ from the given G.P. and then substitute this in the formula ${S_n} = a\left( {\dfrac{{{r^n} - 1}}{{r - 1}}} \right)$

Given, $ \sqrt {17} ,\sqrt {68,} \sqrt {272} ,\sqrt {1088} ....$ $ \\ $ Rewriting the equation as  $ \sqrt {17} ,2\sqrt {17,} 4\sqrt {17} ,8\sqrt {17} \,....$ $ \\ {\text{Formula of sum to first }} n {\text{ terms of the G}}{\text{.P}}{\text{. is:}} \\   \Rightarrow {S_n} = a\left( {\dfrac{{{r^n} - 1}}{{r - 1}}} \right) \\  {\text{Substitute }}a = \sqrt {17} {\text{ and }}r = 2{\text{ in the formula}} \\   \Rightarrow {S_n} = \sqrt {17} \left( {\dfrac{{{2^n} - 1}}{{2 - 1}}} \right)\, \\   \Rightarrow {S_n} = \sqrt {17} \left( {{2^n} - 1} \right) \\  $

 Write fourth term from the end of the G.P. $4, 12, 36, ....... 8748.$

 Write the given G.P. in reverse order and find the first term and common ratio of this G.P. Then substitute values of $a,\,r$ and $n$ in the equation for ${n^{th}}$ term of G.P. 

Given, $4,12,36\,.......8748$ $ \\ $ Rewrite the equation in reverse order, $ 8748, 2916, ....., 12, 4$ $ \\  {\text{Common ratio and first term of this G}}{\text{.P}}{\text{. will be:}} \\   \Rightarrow a = 8748,\,\,r = \dfrac{1}{3} \\  {\text{Formula to find  }}{n^{th}}{\text{ term of a G}}{\text{.P:}} \\   \Rightarrow {a_n} = a{r^{n - 1}}\, \\  {\text{Substitute }}a = 8748,\,r = \dfrac{1}{3}{\text{ and }}n = 4{\text{ in the formula,}} \\   \Rightarrow \,{a_4} = 8748{\left( {\dfrac{1}{3}} \right)^3} \\   \Rightarrow {a_4} = \dfrac{{8748}}{{27}} \\   \Rightarrow {a_4} = 324 \\  {\text{Fourth term from the end of given G}}{\text{.P}}{\text{. is }}324. \\  $

 Fifth term of a G.P. is $\dfrac{1}{{16}}$ and its eighth term is $\dfrac{{ - 1}}{{128}}$. Write the sum of the first four terms of GP.

 Use the given fifth term and eighth term of the G.P. to find $a$and $r$. Then substitute$a$,$n$ and $r$ in formula to find sum of first $n$ term of G.P. 

Given, $a_5 = ar^4 = \dfrac{1}{16} ----(1) \\ a_8 = ar^7 = -\dfrac{1}{128} ----(2) \\ \text{Divide equation }(ii) \text{ by equation }(i) \\ \dfrac{ar^7}{ar^4} = \dfrac{-\dfrac{1}{128}}{\dfrac{1}{16}} = -\dfrac{1}{8} = -\dfrac{1}{2} \\ \text{Substitute } r = -\dfrac{1}{2} \text{ in equation }(i) \\ a\left(-\dfrac{1}{2}\right)^4 = \dfrac{1}{16} = 1 \\ \text{Formula for sum of first } n \text{ terms of G.P is} \\ S_n = a\left(\dfrac{1-r^n}{1-r}\right) \\ \text{Substitute } a = 1, r = -\dfrac{1}{2} \text{ and } n = 4 \text{ in the formula} \\ S_4 = 1\left(\dfrac{1-\left(-\dfrac{1}{2}\right)^4}{1-\left(-\dfrac{1}{2}\right)}\right) = \dfrac{5}{8}$

 ${10^{th}}$ term of a G.P. is $2048$ and its ${6^{th}}$ term is $128$. Write the sum of its ${12^{th}}$ term and ${14^{th}}$ term.

 Use equation to find ${n^{th}}$ term of G.P. and compare the given terms to find  and . Then find ${12^{th}}$ and ${14^{th}}$ terms.

We know that, $ \Rightarrow {a_n} = a{r^{n - 1}}$ We are given, $ &nbsp;&nbsp;&nbsp;\Rightarrow {a_{10}} = a{r^9} = 2048...(i) \\ &nbsp;&nbsp;&nbsp;\Rightarrow {a_6} = a{r^5} = 128...(ii) \\ &nbsp;&nbsp;{\text{Divide equation }}\left( {\text{i}} \right){\text{ by }}\left( {{\text{ii}}} \right) \\ &nbsp;&nbsp;&nbsp;\Rightarrow \dfrac{{a{r^9}}}{{a{r^5}}} = \dfrac{{2048}}{{128}} \\ &nbsp;&nbsp;&nbsp;\Rightarrow {r^4} = 16 \\ &nbsp;&nbsp;&nbsp;\Rightarrow r = 2 \\ &nbsp;$ Substituting the value of r in equation (ii) $ &nbsp;&nbsp;&nbsp;\Rightarrow 32a = 128 \\ &nbsp;&nbsp;&nbsp;\Rightarrow a = 4 \\ &nbsp;$ So, $ &nbsp;&nbsp;&nbsp;\Rightarrow {a_{12}} = a{r^{11}} = 4{(2)^{11}} = 8192 \\ &nbsp;&nbsp;&nbsp;\Rightarrow {a_{14}} = a{r^{13}} = 4{(2)^{13}} = 32768 \\ &nbsp;$ Hence, $ \Rightarrow {a_{12}} + = {a_{14}} = 8192 + 32768 = 40960$