What is the sum of each arithmetic series 50+44+38+….+8?

# What is the sum of each arithmetic series $50+44+38+....+8?$

1. A
242
2. B
232
3. C
252
4. D
262

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### Solution:

The series of arithmetic sequence which is $50,44,38,....8$
The arithmetic expression is expressed as    The first term be $t{1}_{}$and the difference be
Where $d=t2-t1=t3-t2=t4-t3$
The formula is $\mathit{tn}=t1+\left(n-1\right)d$

The general formula for n terms is  S${{n=\frac{n}{2}}_{}=\left[2t1+\left(n-1\right)d\right].}_{}$
If 8 be the term of the series , then $\mathit{tk}=t1+\left(k-1\right)d.$
$\mathit{tk}=t1+\left(k-1\right)d$
$8=50+\left(k-1\right)-6$
$6\left(k-1\right)=50-8=42$
$k=\frac{42}{6}+1=8.$
The sum of the series $S$8$=\frac{8}{2}\left[2×50-6\left(8-1\right)\right]=232.$

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