Question

Sum to n terms of gp series whose rth term is 3r - 1 ​

Answer 1

GIVEN :–

• rth term of G.P. = 3r - 1

TO FIND :–

• Sum of n terms of G.P. = ?

SOLUTION :–

• We know that nth term of G.P. –

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$\bf \implies \large{ \boxed{ \bf T_n = ar^{n-1}}}$

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• According to the question –

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$\bf \implies \bf T_r = 3r - 1$

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• Put r = 1 –

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$\bf \implies \bf T_1 = 3(1) - 1$

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$\bf \implies \bf ar^{1-1} = 3(1) - 1$

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$\bf \implies \bf ar^{0} = 3 - 1$

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$\bf \implies \large{ \boxed{ \bf a = 2}}$

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• Put r = 2 –

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$\bf \implies \bf T_2 = 3(2) - 1$

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$\bf \implies \bf ar^{2-1} = 6 - 1$

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$\bf \implies \bf ar^{1} =5$

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$\bf \implies \bf ar =5$

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$\bf \implies \bf (2)r =5$

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$\bf \implies \large{ \boxed{ \bf r = \dfrac{5}{2}}}$

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• We also know that Sum of n terms –

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$\bf \implies \large{ \boxed{ \bf S_n = \dfrac{a( {r}^{n} - 1)}{r - 1}}}$

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$\bf \implies \bf S_n = \dfrac{2 \left( \left \{ \dfrac{5}{2} \right \}^{n} - 1 \right)}{\left \{ \dfrac{5}{2} \right \} - 1}$

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$\bf \implies \bf S_n = \dfrac{2 \left( \left \{ \dfrac{5}{2} \right \}^{n} - 1 \right)}{ \dfrac{3}{2} }$

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$\bf \implies \bf S_n = \dfrac{4 \left( \left \{ \dfrac{5}{2} \right \}^{n} - 1 \right)}{3}$

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$\bf \implies \large{ \boxed{ \bf S_n =\dfrac{4}{3} { \left( \left \{ \dfrac{5}{2} \right \}^{n} - 1 \right)}}}$

Answer 2

Given ,

• nth term of GP is 3r - 1

Thus ,

First term = 3(1) - 1 = 2

Second term = 3(2) - 1 = 5

$\therefore$ The common ratio of GP will be

r = 5/2

We know that , sum of first n terms of an GP is given by

${\boxed{ \sf{s_{n} = \frac{a \{{(r)}^{n} - 1 \}}{r - 1}} } \: \: \sf{r > 1}}$

Thus ,

$\sf \mapsto s_{n} = \frac{2 \{{( \frac{5}{2} )}^{n} - 1 \}}{ \frac{5}{2} - 1} \\ \\ \sf \mapsto s_{n} = \frac{4 \{ { (\frac{5}{2}) }^{n} - 1 \}}{3}$

Therefore ,

The sum of first n terms of GP is $\sf {s_{n} = \frac{4 \{ { (\frac{5}{2}) }^{n} - 1 \}}{3} }$