Question

A G.P. consists of an even number of terms. If the sum of all the terms is $5$ times the sum of the terms occupying the odd places. Find the common ratio of the G.P.

Answer 1

Hint: The general form of a Geometric progression is
\[{a_1},{a_1}r,{a_1}{r^2},{a_1}{r^3}....{a_1}{r^{n - 1}},{a_1}{r^n}\].
The common ratio is obtained by dividing any term by the preceding term that is $r = \dfrac{{{a_2}}}{{{a_1}}}$.

Complete step-by-step answer:
Let us assume that there will be $2n$ terms in this geometric progression. So the first term will be $'a'$, and the common ratio will be taken as $'r'$.

Now, as it is given in the question itself that
Sum of all the terms $ = 5$ (Sum of the terms occupying the odd places).

According to the above statement, equation will be like,
$ \Rightarrow {a_1} + {a_2} + ... + {a_{2n}} = 5\left( {{a_1} + {a_2} + {a_5} + ... + {a_{2n - 1}}} \right)$

Now put the value of second term$\left( {2n} \right)$ as $\left( {a,r} \right)$, we get
$ \Rightarrow a + ar + ... + a{r^{2n - 1}} = 5\left( {a + a{r^2} + ... + a{r^{2n - 2}}} \right)$……$eq\left( 1 \right)$

Now take the $'a'$ common in the $eq\left( 1 \right)$, we will get
$ \Rightarrow a\left\{ {\dfrac{{1 - {r^{2n}}}}{{1 - r}}} \right\} = 5a\left\{ {{{\dfrac{{1 - \left( {{r^2}} \right)}}{{1 - {r^2}}}}^n}} \right\}$
$
   \Rightarrow 1 + r = 5 \\
   \Rightarrow r = 4 \\
$

Therefore, the common ratio will be 4 for the given G.P.

Note: The behavior of a geometric sequence depends on the value of the common ratio. If the common ratio is,
1. Positive, the terms will all be the same sign as the initial term.
2. Negative, the terms will alternate between positive and negative.
3. Greater than 1, there will be exponential growth towards positive or negative infinity (depending on the sign of the initial term).
4. Between −1 and 1 but not zero, there will be exponential decay towards zero.
5. −1, the progression is an alternating sequence
6. Less than −1, for the absolute values there is exponential growth towards positive and negative infinity (due to the alternating sign).