The last term of an arithmetic series of 20 terms is 195 and common difference is 5. Calculate the sum of the series.
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$\begingroup$The $n$th term of an arithmetic series is given by $$T_n=a+(n-1)d$$where $T_n$ is the $n$th term. If you know $n$, the last term and $d$ you can use this to calculate a value for $a$, the first number. From here you can you the summation formula for an arithmetic series that is $$S_n=\frac{n}{2}\big(2a+(n-1)d\big)$$ where $S_n$ is the sum of the series.
$\endgroup$ $\begingroup$The $nth $ term of an AP is given by $T_n=a+(n-1)d$ where $a$ is the first term and $d$ is the common difference.
Given $a + (20-1)5 = 195$ $\implies a = 100$
The sum of first $n$ terms of an AP is $S_n = \frac{n}{2}(2a + (n-1)d) $
Therefore, $S_{20} = \frac{20}{2}(2\cdot100+(20-1)\cdot5) = 2950$
$\endgroup$ 1 $\begingroup$We know that the formula for an arithmetic series is $$S_n=\frac{n}{2}(a_1+a_n)$$ We already know $a_n$ and $n$, therefore we only have to figure out $a_1$. Is there a way you can use the fact that the common difference is 5 and $n=20$ to figure out $a_1$? (Perhaps, subtract 5 from 195 a specific number of times to reach $a_1$).
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