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$

{\text{a}}{\text{. 1}} \\

{\text{b}}{\text{. }} - 1 \\

{\text{c}}{\text{. 2}} \\

{\text{d}}{\text{. }} - 2 \\

$

Answer

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Hint: - Use common ratio of a G.P $ = \dfrac{{{a_{n + 1}}}}{{{a_n}}}$.

Given geometric series is $ - 1,1, - 1,1...........$

As we know for a G.P say $\left( {{a_{1,}}{a_{2,}}{a_{3,}}{a_{4,}}............{a_n},{a_{n + !}}} \right)$ the common ratio $\left( r \right)$ is written as

$r = \dfrac{{{a_2}}}{{{a_1}}} = \dfrac{{{a_3}}}{{{a_2}}} = \dfrac{{{a_4}}}{{{a_3}}} = ..........\dfrac{{{a_{n + 1}}}}{{{a_n}}}$

Where $\left( {n = 1,2,3..............} \right)$

So, the common ratio $\left( r \right)$for the given series is\[r = \dfrac{1}{{ - 1}} = \dfrac{{ - 1}}{1} = \dfrac{1}{{ - 1}} = - 1\]

Hence option (b) is correct.

Note: - In such types of questions the key concept we have to remember is that always remember the formula of common ratio of a G.P which is stated above, then simplify we will get the required common ratio of a G.P.

Given geometric series is $ - 1,1, - 1,1...........$

As we know for a G.P say $\left( {{a_{1,}}{a_{2,}}{a_{3,}}{a_{4,}}............{a_n},{a_{n + !}}} \right)$ the common ratio $\left( r \right)$ is written as

$r = \dfrac{{{a_2}}}{{{a_1}}} = \dfrac{{{a_3}}}{{{a_2}}} = \dfrac{{{a_4}}}{{{a_3}}} = ..........\dfrac{{{a_{n + 1}}}}{{{a_n}}}$

Where $\left( {n = 1,2,3..............} \right)$

So, the common ratio $\left( r \right)$for the given series is\[r = \dfrac{1}{{ - 1}} = \dfrac{{ - 1}}{1} = \dfrac{1}{{ - 1}} = - 1\]

Hence option (b) is correct.

Note: - In such types of questions the key concept we have to remember is that always remember the formula of common ratio of a G.P which is stated above, then simplify we will get the required common ratio of a G.P.