
Complete the following pattern: $-2,-4,-6,\_\_,\_\_,\_\_$.
Answer
442.5k+ views
Hint: From the given series of arithmetic sequence, we find the general term of the series. We find the formula for ${{t}_{n}}$, the ${{n}^{th}}$ term of the series. From the given sequence we find the common difference between the two consecutive terms. We put the values to get the formula for the general term ${{t}_{n}}$. Then we find three more terms of the series for the solution.
Complete step by step answer:
We have been given a series of arithmetic sequence which is $-2,-4,-6,....$
We express the arithmetic sequence in its general form.
We express the terms as ${{t}_{n}}$, the ${{n}^{th}}$ term of the series.
The first term be ${{t}_{1}}$ and the common difference be $d$ where $d={{t}_{2}}-{{t}_{1}}={{t}_{3}}-{{t}_{2}}={{t}_{4}}-{{t}_{3}}$.
We can express the general term ${{t}_{n}}$ based on the first term and the common difference.
The formula being ${{t}_{n}}={{t}_{1}}+\left( n-1 \right)d$.
The first term is $-2$. So, ${{t}_{1}}=-2$. The common difference is $d={{t}_{2}}-{{t}_{1}}=-4-\left( -2 \right)=-2$.
We express general term ${{t}_{n}}$ as ${{t}_{n}}={{t}_{1}}+\left( n-1 \right)d=-2-2\left( n-1 \right)=-2n$.
Now we need to find three more terms which are ${{t}_{4}},{{t}_{5}},{{t}_{6}}$.
So, putting the values of $4,5,6$ in the equation of ${{t}_{n}}=-2n$, we get
${{t}_{4}}=\left( -2 \right)\times 4=-8,{{t}_{5}}=\left( -2 \right)\times 5=-10,{{t}_{6}}=\left( -2 \right)\times 6=-12$
The next 3 terms in the series $-2,-4,-6,....$ are $-8,-10,-12$.
Note: The sequence is a decreasing sequence where the common difference is a negative number. After nine terms the negative terms of the sequence comes in the series. The common difference will never be calculated according to the difference of greater number from the lesser number.
Complete step by step answer:
We have been given a series of arithmetic sequence which is $-2,-4,-6,....$
We express the arithmetic sequence in its general form.
We express the terms as ${{t}_{n}}$, the ${{n}^{th}}$ term of the series.
The first term be ${{t}_{1}}$ and the common difference be $d$ where $d={{t}_{2}}-{{t}_{1}}={{t}_{3}}-{{t}_{2}}={{t}_{4}}-{{t}_{3}}$.
We can express the general term ${{t}_{n}}$ based on the first term and the common difference.
The formula being ${{t}_{n}}={{t}_{1}}+\left( n-1 \right)d$.
The first term is $-2$. So, ${{t}_{1}}=-2$. The common difference is $d={{t}_{2}}-{{t}_{1}}=-4-\left( -2 \right)=-2$.
We express general term ${{t}_{n}}$ as ${{t}_{n}}={{t}_{1}}+\left( n-1 \right)d=-2-2\left( n-1 \right)=-2n$.
Now we need to find three more terms which are ${{t}_{4}},{{t}_{5}},{{t}_{6}}$.
So, putting the values of $4,5,6$ in the equation of ${{t}_{n}}=-2n$, we get
${{t}_{4}}=\left( -2 \right)\times 4=-8,{{t}_{5}}=\left( -2 \right)\times 5=-10,{{t}_{6}}=\left( -2 \right)\times 6=-12$
The next 3 terms in the series $-2,-4,-6,....$ are $-8,-10,-12$.
Note: The sequence is a decreasing sequence where the common difference is a negative number. After nine terms the negative terms of the sequence comes in the series. The common difference will never be calculated according to the difference of greater number from the lesser number.
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