For a sequence, if Sn = 2 (3^n-1), find the nth term, hence show that the sequence is a G.P.
can someone solve this in an easy way.. I'm hella confused:((
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For a sequence, if Sn = 2 (3^n-1), find the nth term, hence show that the sequence is a G.P.
can someone solve this in an easy way.. I'm hella confused:((
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Given : There is a sequence whose sum of n terms is given as [tex] \sf { S_n = 2(3^n - 1) } [/tex]
[tex] \\ \\ [/tex]
To Find : Find the nth term and Show that the sequence is in G.P
[tex] \\ \\ [/tex]
SolutioN ::
[tex] \bigstar \; {\pmb{\underline{\underline{\textsf{\textbf{ Formula \; Used \; :- }}}}}} [/tex]
[tex] \longrightarrow {\boxed{\sf { T_n = S_n - S_{n - 1} }}} \; \orange\bigstar [/tex]
[tex] \\ [/tex]
Here ::
[tex] \\ \\ [/tex]
[tex] \bigstar \; {\pmb{\underline{\underline{\textsf{\textbf{ Let's \; Calculate \; nth \; term \; :- }}}}}} [/tex]
[tex] \; \; {\longmapsto{\qquad{\sf{ T_n = S_n - S_{n - 1} }}}} \\ \\ \\ \ {\longmapsto{\qquad{\sf{ T_n = 2(3^n - 1) - 2(3^{n-1} - 1) }}}} \\ \\ \\ \ {\longmapsto{\qquad{\sf{ T_n = 2 \bigg[ (3^n - 1) - (3^{n-1} - 1) \bigg] }}}} \\ \\ \\ \ {\longmapsto{\qquad{\sf{ T_n = 2 \bigg[ 3^n - 1 - 3^{n-1} + 1 \bigg] }}}} \\ \\ \\ \ {\longmapsto{\qquad{\sf{ T_n = 2 \bigg[ 3^n - 3^{n-1} \bigg] }}}} \\ \\ \\ \ {\longmapsto{\qquad{\sf{ T_n = 2 \bigg[ 3^n - \dfrac{3^n}{3} \bigg] }}}} \\ \\ \\ \ {\longmapsto{\qquad{\sf{ T_n = 2 \times 3^n \bigg[ 1 - \dfrac{1}{3} \bigg] }}}} \\ \\ \\ \ {\longmapsto{\qquad{\sf{ T_n = 2 \times 3^n \bigg[ \dfrac{3 - 1}{3} \bigg] }}}} \\ \\ \\ \ {\longmapsto{\qquad{\sf{ T_n = 2 \times 3^n \bigg[ \dfrac{2}{3} \bigg] }}}} \\ \\ \\ \ {\longmapsto{\qquad{\sf{ T_n = 4 \times \dfrac{3^n}{3} }}}} \\ \\ \\ \ {\pmb{\longmapsto}{\qquad{\underline{\boxed{\sf{ T_n = 4.3^{n - 1} }}}}}} \; \red\bigstar \\ \\ \\ \ [/tex]
[tex] \\ \\ [/tex]
[tex] \bigstar \; {\pmb{\underline{\underline{\textsf{\textbf{ Checking \; for \; G.P. \; :- }}}}}} [/tex]
[tex] \leadsto [/tex] By using General Term [tex] \sf { Tn = 4.3^{n - 1} } [/tex] we have ::
[tex] \\ [/tex]
Now ; if ::
[tex] \; :\implies \; \sf { \dfrac{a_2}{a_1} = \dfrac{a_3}{a_2} = \dfrac{a_4}{a_3} } \\ \\ [/tex]
[tex] \; :\implies \; \sf { \dfrac{12}{4} = \dfrac{36}{12} = \dfrac{108}{36} } \\ \\ [/tex]
[tex] \; :\implies \; {\underline{\boxed{\sf { r = 3 = 3 = 3 }}}} \; \pink\bigstar \\ \\ [/tex]
[tex] \\ \\ [/tex]
[tex] \; \qquad \; \; \therefore \; [/tex] These Terms are in G.P. !!!
[tex] \\ {\underline{\rule{300pt}{9pt}}} [/tex]
— Hope ; it helps you ! xD