How do you find the sum of each arithmetic series $ 50+44+38+....+8 $?
Answer
576.6k+ views
Hint: From the given series of an arithmetic sequences, 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 difference between two consecutive terms. We put the values to get the formula for the general term $ {{t}_{n}} $ and the formula of summation. Then we put the value in the formula to find the solution.
Complete step by step answer:
We have been given a series of arithmetic sequence which is $ 50,44,38,....,8 $
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 50. So, $ {{t}_{1}}=50 $ . The common difference is $ d={{t}_{2}}-{{t}_{1}}=44-50=6 $ .
We express general term $ {{t}_{n}} $ as $ {{t}_{n}}={{t}_{1}}+\left( n-1 \right)d=50-6\left( n-1 \right)=56-6n $ .
Now we need to find the formula of summation of the arithmetic sequence.
The general formula for n terms is \[{{S}_{n}}=\dfrac{n}{2}\left[ 2{{t}_{1}}+\left( n-1 \right)d \right]\].
We need to find the number of terms in the series $ 50,44,38,....,8 $.
If 8 be the $ {{k}^{th}} $ term of the series, then $ {{t}_{k}}={{t}_{1}}+\left( k-1 \right)d $ . Solving the equation, we get
$ \begin{align}
& {{t}_{k}}={{t}_{1}}+\left( k-1 \right)d \\
& \Rightarrow 8=50+\left( k-1 \right)\left( -6 \right) \\
& \Rightarrow 6\left( k-1 \right)=50-8=42 \\
& \Rightarrow k=\dfrac{42}{6}+1=8 \\
\end{align} $
There are 8 terms in the series $ 50,44,38,....,8 $
The sum of the series will be \[{{S}_{8}}=\dfrac{8}{2}\left[ 2\times 50-6\left( 8-1 \right) \right]=232\]
Therefore, the sum of each arithmetic series $ 50+44+38+....+8 $ is 232.
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 the greater number from the lesser number.
Complete step by step answer:
We have been given a series of arithmetic sequence which is $ 50,44,38,....,8 $
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 50. So, $ {{t}_{1}}=50 $ . The common difference is $ d={{t}_{2}}-{{t}_{1}}=44-50=6 $ .
We express general term $ {{t}_{n}} $ as $ {{t}_{n}}={{t}_{1}}+\left( n-1 \right)d=50-6\left( n-1 \right)=56-6n $ .
Now we need to find the formula of summation of the arithmetic sequence.
The general formula for n terms is \[{{S}_{n}}=\dfrac{n}{2}\left[ 2{{t}_{1}}+\left( n-1 \right)d \right]\].
We need to find the number of terms in the series $ 50,44,38,....,8 $.
If 8 be the $ {{k}^{th}} $ term of the series, then $ {{t}_{k}}={{t}_{1}}+\left( k-1 \right)d $ . Solving the equation, we get
$ \begin{align}
& {{t}_{k}}={{t}_{1}}+\left( k-1 \right)d \\
& \Rightarrow 8=50+\left( k-1 \right)\left( -6 \right) \\
& \Rightarrow 6\left( k-1 \right)=50-8=42 \\
& \Rightarrow k=\dfrac{42}{6}+1=8 \\
\end{align} $
There are 8 terms in the series $ 50,44,38,....,8 $
The sum of the series will be \[{{S}_{8}}=\dfrac{8}{2}\left[ 2\times 50-6\left( 8-1 \right) \right]=232\]
Therefore, the sum of each arithmetic series $ 50+44+38+....+8 $ is 232.
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 the greater number from the lesser number.
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