Question

Find the 3rd term from the end of the GP 2/27, 2/9, 2/3,..... 162.

Answer 1

$Given \: G.P : \frac{2}{27}, \frac{2}{9},\frac{2}{3},\ldots 162$

$First \:term (a) = a_{1} = \frac{2}{27}$

$Common \:ratio (r) = \frac{a_{2}}{a_{1}}$

$= \frac{\frac{2}{9}}{\frac{2}{27}}$

$= \frac{2}{9} \times \frac{27}{2}$

$= 3$

/* Rearranging the sequence in reverse order , we get */

$First \:term = 162$

$Common \:ratio (r) = \frac{1}{3}$

/* Now, new G.P */

$162 , 54, \cdots , \frac{2}{3}, \frac{2}{9},\frac{2}{27}$

$3^{rd} \:term = ar^{2}$

$\boxed {\pink {\because n^{th}\:term (a_{n}= ar^{n-1} }}$

$a_{3} = 162 \times \Big(\frac{1}{3}\Big)^{2}$

$= 162\times \frac{1}{9}$

$= 18$

Therefore.,

$\red{ The \: 3^{rd} \:term \: from \:the \:end }$

$\red{of \: GP } \green { = 18}$

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Answer 2

Answer:

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