Question

Find the first four terms of an Arithmetic Progression whose first term is $-2$ and common difference is $-2.$

Answer 1

Hint: An arithmetic progression has a common difference. If the differences of the consecutive terms are the same, then it is called the common difference of the arithmetic progression. We can find the ${{n}^{th}}$ term of the arithmetic progression by ${{a}_{n}}={{a}_{1}}+\left( n-1 \right)d$ where ${{a}_{n}}$ is the ${{n}^{th}}$ term, ${{a}_{1}}$ is the first term and $d$ is the common difference.

Complete step by step answer:
We have already learnt that an arithmetic progression is a sequence with the differences of the consecutive terms are the same. And this difference is called the common difference.
Let us suppose that we are given with the first term ${{a}_{1}}$ of an arithmetic progression. Also, the common difference $d$ is also given. Then it is easy to find the ${{n}^{th}}$ term ${{a}_{n}}$ by adding the first term to the product of the common difference and $n-1.$
That is, ${{a}_{n}}={{a}_{1}}+\left( n-1 \right)d.$
We can apply this to find the first four terms of the A.P whose first term is $-2$ and common difference is $-2.$
So, ${{a}_{1}}=-2$ and $d=-2.$
Let us find the second term using the above formula. So, here $n-1=1.$
Therefore, ${{a}_{2}}=-2+1\times \left( -2 \right)=-2+-2=-4.$
So, we can find the third term similarly by taking $n-1=2.$
We will get ${{a}_{3}}=-2+2\times \left( -2 \right)=-2-4=-6.$
Now, the fourth term can be found when we fix $n-1=3.$
So, ${{a}_{4}}=-2+3\times \left( -2 \right)=-2-6=-8.$

Hence, the first four terms are $-2,-4,-6,-8.$

Note: This can be directly done by adding the common difference $d$ to the first term, the second term and the third term. Since the first term is ${{a}_{1}}=-2$ and $d=-2,$ the second term is ${{a}_{2}}=-2+\left( -2 \right)=-4.$ The third term is ${{a}_{3}}=-4+\left( -2 \right)=-6$ and the fourth term is ${{a}_{4}}=-6+\left( -2 \right)=-8.$