Question

Find the sum of the given arithmetic series.

24 + 48 + 72 + 96 + ... + 480

Answer 1

\(\normalsize\color{blue}{ a_n=a_1+d(n-1) }\) and in this case, \(\normalsize\color{blue}{ 480=24+24(n-1) }\)
\(\normalsize\color{blue}{ 480=24(1)+24(n-1) }\)
\(\normalsize\color{blue}{ 480=24(n-1+1) }\)
\(\normalsize\color{blue}{ 480=24(n) }\)
\(\normalsize\color{blue}{ n= 20 }\)

Answer 2

So, we found that there are 20 terms.

Answer 3

\(\LARGE\color{blue}{ S_n=\frac{1}{2}~(a_1+a_n)\times n }\)

Answer 4

\(\LARGE\color{blue}{ S_{20}=\frac{1}{2}~(24+480)\times 20 }\)


\(\LARGE\color{blue}{ S_{20}=(24+480)\times 10 }\)


\(\LARGE\color{blue}{ S_{20}=(504)\times 10 }\)


\(\LARGE\color{blue}{ S_{20}=5040 }\)

Answer 5

So the answer is 5040.

Answer 6

thank you! :)

Answer 7

find the sum of the given arithmetic series. (:
1.30+23+16+9+...(-30)
2. 26+19+12+5+...+(-37)
3.27+20+13+6+...+(-36)

Answer 8

can you help me on these also? :)