The sum of the first 15 terms of an arithmetic series is 1290. The sum of the first 16 terms is 1464. Find t16. Find S20

sum(15) = (15/2)(2a + 14d) = 1290

2a + 14d =172

sum(16) = (16/2)(2a + 15d) = 1464
2a + 15d = 183
subtract them:
d = 11
sub into 2a + 14d = 172
2a + 154 = 172
2a = 18
a = 9

term(16) = a + 15d = 174

sum(20) = 10(2a + 19d) = 10(18+209) = 2270

BTW, we could have found term (16)
by Sum(16) - sum(15) = 1464-1290 = 174

but we needed both a and d to find sum(20)

To find the value of t16 (the 16th term of the arithmetic series), we will use the formula for the sum of an arithmetic series:

\[S_n = \frac{n}{2}(2a + (n-1)d)\]

where \(S_n\) is the sum of the first n terms, a is the first term, d is the common difference, and n is the number of terms.

Given that the sum of the first 15 terms is 1290, we can substitute these values into the formula:

\[1290 = \frac{15}{2}(2a + (15-1)d)\]

Simplifying this equation gives us:

\[1290 = \frac{15}{2}(2a + 14d)\]

Divide both sides of the equation by 15/2 (or multiply by 2/15) to isolate the expression inside the parentheses:

\[86 = 2a + 14d\]

Similarly, for the sum of the first 16 terms being 1464, we have:

\[1464 = \frac{16}{2}(2a + (16-1)d)\]
\[1464 = 8(2a + 15d)\]
\[183 = 2a + 15d\]

Now we have a system of two equations:

\[
\begin{align*}
86 &= 2a + 14d \\
183 &= 2a + 15d \\
\end{align*}
\]

We can solve this system of equations to find the values of a and d.

Subtract the first equation from the second equation to eliminate "a":

\[183 - 86 = (2a + 15d) - (2a + 14d)\]
\[97 = d\]

Now that we know the value of d, we can substitute it back into either of the original equations to solve for a.

Substituting d = 97 into the first equation:

\[86 = 2a + 14(97)\]
\[86 = 2a + 1358\]
\[2a = 86 - 1358\]
\[2a = -1272\]
\[a = -636\]

So, the first term (a) is -636 and the common difference (d) is 97.

Now, we can find t16 (the 16th term) by using the formula for the nth term of an arithmetic series:

\[t_n = a + (n-1)d\]

Substituting the values we found:

\[t_{16} = -636 + (16 - 1) \cdot 97\]

Simplifying:

\[t_{16} = -636 + 15 \cdot 97\]

\[t_{16} = -636 + 1455\]

\[t_{16} = 819\]

Therefore, the 16th term (t16) of the arithmetic series is 819.

To find S20 (the sum of the first 20 terms), we will use the formula for the sum of an arithmetic series:

\[S_n = \frac{n}{2}(2a + (n-1)d)\]

Substituting the values we found previously:

\[S_{20} = \frac{20}{2}(2 \cdot -636 + (20-1) \cdot 97)\]

Simplifying:

\[S_{20} = 10(-1272 + 19 \cdot 97)\]

\[S_{20} = 10(-1272 + 1843)\]

\[S_{20} = 10(571)\]

\[S_{20} = 5710\]

Therefore, the sum of the first 20 terms (S20) of the arithmetic series is 5710.