Given,
The arithmetic sequence is 3,9, 15, ........
Required
The sum of 34 terms of the series.
Here,
The first term of the series is,
a = 3
The common difference of the series is,
d = 9 -3 = 6
The number of terms are 34
The sum of the series is calculated as,
[tex]\begin{gathered} Sum=\frac{n}{2}(2a+(n-1)d) \\ =\frac{34}{2}(2\times3+(34-1)6) \\ =17(6+33\times6) \\ =3468 \end{gathered}[/tex]Hence, the sum of the series is 3468.
