Question

How do you find the common ratio for \[64,-32,16,-8,4.....\]?

Answer 1

Hint: In this problem, we have to find the common ratio for the given geometric series. We know that, in a geometric series, the common ratio is the ratio of a term with respect to its preceding term. We can divide the term with respect to its preceding term one by one and check whether the result is the same for every term to find the common ratio.

Complete step by step solution:
We know that the geometric series given for which we have to find the common ratio is,
\[64,-32,16,-8,4.....\]
We can assume that, let the ratio between successive terms \[{{a}_{n}}\] and \[{{a}_{n+1}}\] be,
\[{{r}_{n}}=\dfrac{{{a}_{n+1}}}{{{a}_{n}}}\]
We know that the given terms are,
 \[64,-32,16,-8,4.....\]
Where,
\[{{a}_{1}}=64,{{a}_{2}}=-32,{{a}_{3}}=16,{{a}_{4}}=-8,{{a}_{5}}=4\]
We can now find the value of \[{{r}_{1}}\],
 \[\Rightarrow {{r}_{1}}=\dfrac{{{a}_{2}}}{{{a}_{1}}}=\dfrac{-32}{64}=-\dfrac{1}{2}\] …… (1)
We can now find the value of \[{{r}_{2}}\],
\[\Rightarrow {{r}_{2}}=\dfrac{{{a}_{3}}}{{{a}_{2}}}=\dfrac{16}{-32}=-\dfrac{1}{2}\] ……. (2)
We can now find the value of \[{{r}_{3}}\],
 \[\Rightarrow {{r}_{3}}=\dfrac{{{a}_{4}}}{{{a}_{3}}}=\dfrac{-8}{16}=-\dfrac{1}{2}\] …….. (3)
We can now find the value of \[{{r}_{4}}\],
\[\Rightarrow {{r}_{4}}=\dfrac{{{a}_{5}}}{{{a}_{4}}}=\dfrac{4}{-8}=-\dfrac{1}{2}\] ……… (4)
We can now compare the values of (1), (2), (3), (4), we have the same values.
Therefore, the common ratio for the given geometric series \[64,-32,16,-8,4.....\] is \[-\dfrac{1}{2}\].

Note: Students make mistakes while finding the value of common ration by substituting the correct values in the ratio. We should remember that we can divide the term with respect to its preceding term one by one and check whether the result is the same for every term to find the common ratio.