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Sequences Quiz 2

mathematics

 If the $n^{th}$ terms of G.P is $3072$ and ${4^{th}}$ term of G.P is $24$. Find the value of ${n^{th}}$ term if the first term is one more than its common ratio.

 Use ${4^{th}}$ term and compare the value for getting the value of $'a'$ and $'r'$ and then apply the ratio formula.

  $  {\text{Consider the given }}{{\text{4}}^{th}}{\text{ term of G}}{\text{.P = 24}}   \\  \therefore {\text{a}}{{\text{r}}^3} = 3 \times {2^3}{\text{ on comparing}}   \\  \therefore {\text{a = 3}}   \\  {\text{b = 2}}   \\  {\text{a}}{{\text{r}}^{n - 1}} = 3072   \\  3 \times {\left( 2 \right)^{n - 1}} = 3072   \\  {\left( 2 \right)^{n - 1}} = 1024   \\  {\left( 2 \right)^{n - 1}} = {\left( 2 \right)^{10}}   \\  \therefore n = 11   \\  $

 Find the sum of $n$ terms of A.P, if ${q^{th}}$ term is $3{q^2} + q$.

 Use ${q^{th}}$ term to find the first term and common difference and then solve. 

  $  {\text{Given }}{{\text{q}}^{{\text{th}}}}{\text{ term is 3}}{{\text{q}}^2} + q   \\  \therefore {\text{First term}} = 3 + 1 = 4   \\  {\text{Second term }} = 12 + 2 = 14   \\  \therefore {\text{ Common difference }} = 14 - 4 = 10   \\  {\text{Sum of n terms}}   \\   \Rightarrow {S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]   \\   \Rightarrow {S_n} = \dfrac{n}{2}\left[ {2 \times 4 + \left( {n - 1} \right)10} \right]   \\   \Rightarrow {S_n} = \dfrac{n}{2}\left( {10n - 2} \right)   \\   \Rightarrow {S_n} = 5{n^2} - n   \\  $

 If the ratio of ${3^{rd}}$ term and ${7^{th}}$ term is $\dfrac{2}{3}$, common difference is $1$ more from first term. Find the sum of $n$ terms.

 Use ratio to find the value of $'a'$ and $'d'$ and further solve for getting ${S_n}$. 

$  {\text{Given ratio }}\dfrac{{{3^{rd}}term}}{{{7^{th}}term}} = \dfrac{2}{3}   \\  \therefore \dfrac{{a + 2d}}{{a + 6d}} = \dfrac{2}{3}   \\   \Rightarrow a = \dfrac{{ - 6}}{5}   \\  \therefore d = \dfrac{{ - 6}}{5} + 1   \\  d = \dfrac{{ - 1}}{5}   \\  \therefore {\text{sum of n terms is  = }}\dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right)   \\   \Rightarrow \dfrac{n}{2}\left( {\dfrac{{ - 12}}{5} + \left( {n - 1} \right) - \dfrac{1}{5}} \right)   \\   \Rightarrow \dfrac{n}{2}\left( {\dfrac{{ - 12}}{5} + \dfrac{1}{5} - \dfrac{n}{5}} \right)   \\   \Rightarrow \dfrac{{n\left( { - 11 - n} \right)}}{{2 \times 5}}   \\   \Rightarrow \dfrac{{ - n\left( {n + 11} \right)}}{{10}}   \\  $

 Find the sum of $n$ terms of A.P. If ${7^{th}}$ term of A.P is $2$ times of ${3^{rd}}$ term and sum of ${9^{th}}$ and ${5^{th}}$ term is $144$.

 Apply elimination rule to find the value of ‘a’ and ‘d’ and later find sum of n terms. 

 $  Given{\text{ }}{{\text{7}}^{th}}{\text{ term 2 times of }}{{\text{3}}^{rd}}{\text{ term}}   \\  \therefore {\text{a + 6d = 2}}\left( {a + 2d} \right)   \\   \Rightarrow a - 2d = 0..........\left( i \right)   \\  {9^{th}}{\text{ term  + }}{{\text{5}}^{th}}{\text{ term  =  144}}   \\  {\text{a  +  8d  +  a  +  4d  =  144}}   \\  {\text{a  +  6d  =  72}}.........\left( {ii} \right)   \\  from{\text{ }}\left( i \right){\text{ and }}\left( {ii} \right){\text{ equation by using elimination method}}{\text{.}}   \\  {\text{a  =  18}}   \\  {\text{d  =  9}}   \\  {\text{sum of n term }}{{\text{S}}_n} = \dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right)   \\   \Rightarrow \dfrac{n}{2}\left( {2 \times 18 + 9n - 9} \right)   \\   \Rightarrow \dfrac{{9n\left( {n + 3} \right)}}{2}   \\  $

 Group of $10$ people engaged to finish a work in a certain number of days. $2$ people add more and on the second day more $2$ people are added and next day and soon. It took $2$ days less to finish this work. Find the number of days that work can be finished. 

 Use given information to make sequences to determine value of first term and common difference and then apply the sum of n terms of an A.P formula. 

 $  {\text{Let }}x{\text{ be the number of days in which }}10{\text{ peoples finished the work}}   \\  {\text{According to question}}   \\  10x = 10 + 14 + 18 + ....... + \left( {x - 2} \right)   \\  \therefore {\text{Series }}10 + 14 + 18 + .......... + \left( {x - 2} \right)   \\  {\text{First term(a)}} = 10   \\  {\text{Common difference}}\left( d \right) = 2   \\  {\text{Number of terms}}\left( n \right) = x - 2   \\  {\text{Apply }}{{\text{S}}_{\text{n}}} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]   \\  \therefore 10x = \left( {\dfrac{{x - 2}}{2}} \right)\left[ {20 + \left( {x - 3} \right)2} \right]   \\   \Rightarrow 20x = \left( {x - 2} \right)\left( {20 + 2x - 6} \right)   \\   \Rightarrow {x^2} + 5x - 24 = 0   \\   \Rightarrow \left( {x + 8} \right)\left( {x - 3} \right) = 0   \\  \therefore x =  - 8\left( {{\text{Not possible}}} \right),x = 3   \\  \therefore {\text{Total number of days}} = 3 - 2 = 1   \\  $