Question

If $a=-4$ and $r=-2$, find the GP. \[\]

Answer 1

Hint:We use the fact that that a finite geometric progression or GP sequence in terms of first term $a$ and common ratio $r$ for $n$ terms can be written as $a,a{{r}^{2}},a{{r}^{3}},a{{r}^{4}},...,a{{r}^{n-1}}$. We see that the given question does not mention the number of terms , so it will be infinite GP as $a,a{{r}^{2}},a{{r}^{3}},a{{r}^{4}},...$\[\]

Complete step by step answer:
A sequence is defined as the enumerated collection of numbers where repetitions are allowed and the order of the numbers matters. It can also be expressed as a one-one map from the natural numbers set to real numbers. The members of the sequence are called terms. Mathematically, a sequence with infinite terms is written as
\[\left( {{x}_{n}} \right)={{x}_{1}},{{x}_{2}},{{x}_{3}},...\]
If the sequence has finite terms terminated by a term then we write the sequence as
\[\left( {{x}_{n}} \right)={{x}_{1}},{{x}_{2}},{{x}_{3}},...{{x}_{n}}\]
Geometric sequence otherwise known as Geometric progression , abbreviate d as GP is a type sequence where the ratio between any two consecutive numbers is constant . If $\left( {{x}_{n}} \right)={{x}_{1}},{{x}_{2}},{{x}_{3}},...$ is an GP, then
\[\dfrac{{{x}_{2}}}{{{x}_{1}}}=\dfrac{{{x}_{3}}}{{{x}_{1}}}...=r...(1)\]
Here ratio between two terms is called common ratio and denoted as $r$. The first term is conventionally denoted as $a$. We put ${{x}_{1}}=a$ in equation (1) and have,
\[r=\dfrac{{{x}_{2}}}{{{x}_{1}}}=\dfrac{{{x}_{2}}}{a}\Rightarrow {{x}_{2}}=ar\]
We can similarly put ${{x}_{2}}=ar$ in equation (1) and have,
\[r=\dfrac{{{x}_{3}}}{{{x}_{2}}}=\dfrac{{{x}_{3}}}{ar}\Rightarrow {{x}_{3}}=ar\times r=a{{r}^{2}}\]
We can similarly get ${{x}_{4}}=a{{r}^{3}},{{x}_{5}}=a{{r}^{4}},...$ . We observe that the ${{n}^{\text{th}}}$ term of GP sequence is $a{{r}^{n-1}}$ and we can write the infinite GP sequence in terms of first term $a$ and common ratio $r$ as
\[a,a{{r}^{2}},a{{r}^{3}},a{{r}^{4}},...\]
We are given in the question that $a=-4$ and $r=-2$ which according to conventional notations the first term is $-4$and common ratio is $-2.$So we have the second term of the GP ${{x}_{2}}=ar=\left( -4 \right)\left( -2 \right)=8$, the third term of the GP as ${{x}_{3}}=a{{r}^{2}}=\left( -4 \right){{\left( -2 \right)}^{2}}=-16$, the fourth term of the GP as ${{x}_{4}}=a{{r}^{3}}=\left( -4 \right){{\left( -2 \right)}^{3}}=32$ and so on. So we can write the GP as,
\[-4,8,-16,32,...\]

Note:
 We must be careful of the confusion of GP from AP otherwise known as arithmetic sequence where there is a common difference between any two consecutive terms not common ratio. We can also write a GP considering $a$ as the last term in the form of $...,\dfrac{a}{{{r}^{2}}},\dfrac{a}{r},a$. The sum of first $n$ terms of a GP is given by the formula $\dfrac{a\left( 1-{{r}^{n}} \right)}{1-{{r}^{n}}}$