Question

Find the sum of the first 17 terms of the arithmetic sequence 10, 14, 18, 22, 26...

Answer 1

10, 14, 18 ....

notice, we get the next term by simply adding 4 to the current term, thus "4" is the "common difference, and we know that 10 is the first term.


`\bf n^(th)\textit{ term of an arithmetic sequence} \\\\ a_n=a_1+(n-1)d\qquad \begin{cases} n=n^(th)\ term\\ a_1=\textit{first term's value}\\ d=\textit{common difference}\\ ----------\\ a_1=10\\ d=4\\ n=17 \end{cases} \\\\\\ a_(17)=10+(17-1)(4)\implies a_(17)=10+(16)(4) \\\\\\ a_(17)=10+64\implies a_(17)=74\\\\ -------------------------------`


`\bf \textit{ sum of a finite arithmetic sequence} \\\\ S_n=\cfrac{n(a_1+a_n)}{2}\qquad \begin{cases} n=n^(th)\ term\\ a_1=\textit{first term's value}\\ ----------\\ a_1=10\\ a_(17)=74\\ n=17 \end{cases} \\\\\\ S_(17)=\cfrac{17(10+74)}{2}\implies S_(17)=\cfrac{17(84)}{2}\implies S_(17)=714`

Answer 2

The answer would be 714.

Common difference =4An=A1+(n-1)dA17=10+(17-1)(4)=10+(16)(4)=10+64=74
Sn=n(a1+an)/2S17=17(10+74)/2=714