Question

How do you find the $ {{12}^{th}} $ term of the arithmetic sequence $ 20,14,8,2,-4,..... $ ?

Answer 1

Hint: From the given series of an 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 put the value of $ n=12 $ to find the solution.

Complete step by step answer:
We have been given a series of arithmetic sequences which is $ 20,14,8,2,-4,..... $
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 be and the common difference.
The formula being $ {{t}_{n}}={{t}_{1}}+\left( n-1 \right)d $ .
The first term is 20. So, $ {{t}_{1}}=20 $ . The common difference is $ d={{t}_{2}}-{{t}_{1}}=14-20=6 $ .
We express general term $ {{t}_{n}} $ as $ {{t}_{n}}={{t}_{1}}+\left( n-1 \right)d=20-6\left( n-1 \right)=26-6n $ .
Now we need to find the $ {{12}^{th}} $ term of the arithmetic sequence which is $ {{t}_{12}} $ .
We put the value of $ n=12 $ in the formula of $ {{t}_{n}}=26-6n $ .
So, $ {{t}_{n}}=26-6n=26-6\times 12=26-72=-46 $ .
Therefore, the $ {{12}^{th}} $ term of the arithmetic sequence $ 20,14,8,2,-4,..... $ is $ -46 $ .

Note:
The sequence is a decreasing sequence where the common difference is a negative number. After the four terms, the negative terms of the sequence comes in the series. The common difference will never be calculated according to the difference of the greater number from the lesser number.