Geometric Progression - Practice Questions & MCQ

A geometric sequence is a sequence where the first term is non-zero and the ratio between consecutive terms is always constant. The 'constant factor' is called the common ratio and is denoted by ' $r$ '. $r$ is also a non-zero number.

The first term of a G.P. is usually denoted by 'a'.
If $a_1, a_2, a_3 \ldots . a_{n-1}, a_n$ is in geometric progression
then, $r=\frac{a_2}{a_1}=\frac{a_3}{a_2}=\ldots=\frac{a_n}{a_{n-1}}$
Eg,
- $2,6,18,54, \ldots . .(a=2, r=3)$
- $4,2,1,1 / 2,1 / 4, \ldots . .(a=4, r=1 / 2)$
- $-5,5,-5,5, \ldots \ldots(a=-5, r=-1)$

General Term of a GP
If ' $a$ is the first term and ' $r$ ' is the common ratio, then.

$$
\begin{aligned}
& a_1=a=a r^{1-1} \quad\left(1^{\text {st }} \text { term }\right) \\
& a_2=a r=a r^{2-1}\left(2^{\text {nd }} \text { term }\right) \\
& a_3=a r^2=a r^{3-1}\left(3^{\text {rd }} \text { term }\right) \\
& \cdots \\
& \cdots \\
& a_n=a r^{n-1}\left(\mathrm{n}^{\text {th }} \text { term }\right)
\end{aligned}
$$

So, the general term or $\mathrm{n}^{\text {th }}$ term of a geometric progression is $a_n=a r^{n-1}$a

Increasing and Decreasing GP
For a GP to be increasing or decreasing, $r>0$. If $r<0$, then the terms of G.P. are alternately positive and negative so neither increasing nor decreasing.
\begin{tabular}{||c||c||c||c||c||}
\hline$a$ & $a>0$ & $a>0$ & $a<0$ & $a<0$ \\
\hline$r$ & $r>1$ & $0<r<1$ & $r>1$ & $0<r<1$ \\
\hline Result & Increasing & Decreasing & Decreasing & Increasing \\
\hline
\end{tabular}

Important Properties of a GP

If $a, b, c$ are in GP, then $b^2=a . c$
If each term of a G.P. is multiplied by a fixed constant or divided by a non-zero fixed constant then the resulting series is also in G.P. with the same common ratio as the original series.

If each term of a G.P. is raised to some real number $m$, then the resulting series is also in G.P.

Important Properties of a GP
4. If $a_1, a_2, a_3 \ldots$ and $b_1, b_2, b_3 \ldots$ are two G.P.'s, then $a_1 b_1, a_2 b_2, a_3 b_3 \ldots$ and $\frac{a_1}{b_1}, \frac{a_2}{b_2}, \frac{a_3}{b_3} \ldots$ are also G.P.s.
5. If $a_1, a_2, a_3, \ldots ., a_{n-1}, a_n$ are in G.P. with common ratio r , then $\log a_1, \log a_2, \log a_3 \ldots \ldots$.is in A.P. and the converse also holds true.
6. If three numbers in G.P. whose product is given are to be taken, then take them as $\mathrm{a} / \mathrm{r}$, a, ar.
7. If four numbers in G.P. whose product is given are to be taken, then take them as $\frac{a}{r^3}, \frac{a}{r}, a r, a r^3$.
8. The product of terms equidistant from the start and end of the G.P. is constant and it equals product of the first and the last terms.

Some questions based on Geometric Progression

Q: If the product of three terms of a GP is 512, and the sum of their product in pair is 224, then the sum of three numbers will be

Sol: It is given that three terms are in G.P. and their product is 512,

So take three numbers as $\mathrm{a} / \mathrm{r}$, a, ar
Then, the product, $\mathrm{a} / \mathrm{r} \cdot \mathrm{a} \cdot \mathrm{ar}=512 \Rightarrow \mathrm{a}^3=512 \Rightarrow \mathrm{a}=8$
Now,
The sum of the product in the pair is 224
So, $a / r \cdot a+a \cdot a r+a / r \cdot a r=224$

$
\Rightarrow a^2(1 / r+r+1)=224
$
Put the value of $a$,

$
\begin{aligned}
& \Rightarrow 64(1 / r+r+1)=224 \\
& \Rightarrow(1 / r+r+1)=7 / 2 \\
& \Rightarrow r=2 \text { or } r=1 / 2
\end{aligned}
$
So, the three numbers which are in GP are $4,8,16$ or $16,8,4$
Sum is $16+8+4=28$