Given that in an Arithmetic series,t2=7 and t5=19,find s13 ?

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embizze | High School Teacher | (Level 1) Educator Emeritus

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An arithmetic series is a series whose terms differ by a constant,d.

To find the finite sum of an arithmetic series we use:

`S_n=n/2(t_1+t_n)` where `S_n` is the sum of the first n terms, `t_1` is the first term and `t_n` the `n^(th)` term.

The formula for the `n^(th)` term of an arithmetic sequence is:

`t_n=t_1+(n-1)d`

We can use the given terms, `t_2=7` and `t_5=19` to find `t_1,d` .

`7=t_1+(2-1)d==> 7=t_1+d`

`19=t_1+(5-1)d==>19=t_1+4d`

Subtracting the first from the second yields:

`12=3d==>d=4`

If d=4, then `t_1=3`

**Note that this generates the sequence 3,7,11,15,19,... with `t_2=7,t_5=19` as required.**

So `S_13=13/2(3+t_13)` , but `t_13=3+(13-1)4==>t_13=51`

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`S_13=13/2(3+51)=13(27)=351`

The sum of the first 13 terms is 351.

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