Question

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Answer 1

Hope you find this Solution Helpful ☺️

Answer 2

Question

Find ths sum of the first 6 terms of the sequence 3645, 1215, 405.

$\large\underline{\sf{Solution-}}$

Given series is

$\rm \: 3645, \: 1215, \: 405, \: - - -$

Let we first check the series

We have given

$\rm \: a_1 = 3645$

$\rm \: a_2 = 1215$

$\rm \: a_3 = 405$

Consider,

$\rm \: \dfrac{a_2}{a_1} = \dfrac{1215}{3645} = \dfrac{1}{3}$

Consider,

$\rm \: \dfrac{a_3}{a_2} = \dfrac{405}{1215} = \dfrac{1}{3}$

$\rm\implies \:\dfrac{a_2}{a_1} = \dfrac{a_3}{a_2}$

This implies, the given series forms a GP series.

So, we have

$\rm \: a_1 = 3645$

$\rm \: r = \dfrac{1}{3}$

$\rm \: n = 6$

We know,

↝ Sum of n  terms of an geometric series is,

$\begin{gathered}\red\bigstar\:\:{\underline{\orange{\boxed{\bf{\green{S_n\:=\dfrac{a \: (1 - {r}^{n} )}{1 - r} \: \: \sf \: provided \: that \: r \: \ne \: 1}}}}}} \\ \end{gathered}$

Wʜᴇʀᴇ,

Sₙ is the sum of n terms of GP..

a is the first term of the sequence.

n is the no. of terms.

r is the common ratio.

So, on substituting the values, we get

$\rm \: S_6 = 3645 \times \dfrac{1 - {\bigg(\dfrac{1}{3} \bigg) }^{6} }{1 - \dfrac{1}{3} }$

$\rm \: S_6 = 3645 \times \dfrac{1 - \dfrac{1}{729} }{\dfrac{3 - 1}{3} }$

$\rm \: S_6 = 3645 \times \dfrac{\dfrac{729 - 1}{729} }{\dfrac{2}{3} }$

$\rm \: S_6 = 3645 \times \dfrac{728}{729} \times \dfrac{3}{2}$

$\rm \: S_6 = 3645 \times \dfrac{364}{243}$

$\rm \: S_6 = 15 \times 364$

$\bf\implies \:S_6 = 5460$

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↝ nᵗʰ term of an geometric series is,

$\begin{gathered}\red\bigstar\:\:{\underline{\orange{\boxed{\bf{\green{a_n\:=\: {a \: r}^{n \: - \: 1} }}}}}} \\ \end{gathered}$

Wʜᴇʀᴇ,

aₙ is the nᵗʰ term.

a is the first term of the sequence.

n is the no. of terms.

r is the common ratio.