Question

How do you find the sum of the arithmetic sequence given ${a_1} = 81$, ${a_n} = - 13$ and $n = 20$?

Answer 1

We can solve this using the formula of sum of all terms in a finite arithmetic progression with the last term given. Since we have the first and last term we use the formula,${S_n} = \dfrac{n}{2}({a_1} + {a_n})$. Using this we can solve this. (If the last term is not given we use the sum of infinite series formulas and we need the common difference value ‘d’.)

Complete step by step answer:
Given,
${a_1} = 81$, ${a_{20}} = {a_n} = - 13$ and $n = 20$.
We know the formula of sum of all terms in a finite arithmetic progression is
${S_n} = \dfrac{n}{2}({a_1} + {a_n})$, where ${a_1}$ is first term, ${a_n}$ is the last term and ‘n’ is number of terms.
Substituting we have,
${S_{20}} = \dfrac{{20}}{2}(81 - 13)$
${S_{20}} = 10 \times 68$
$\Rightarrow {S_{20}} = 680$

Hence the sum of the arithmetic sequence is $\Rightarrow {S_{20}} = 680$. That is the sum of the first 20 term

Note: Remember all the formulas. We also know the sum of infinite terms formula in A.P is ${S_n} = \dfrac{n}{2}(2a + (n - 1)d)$, where ‘d’ is the common difference. In arithmetic progression we find common differences but in geometric progression we find common ratios. We also know the general form of A.P is $a,a + d,a + 2d,a + 3d...........,a + (n - 1)d$. The sum of infinite terms in geometric progression is ${S_n} = \dfrac{{a({r^n} - 1)}}{{(r - 1)}}$. Where ‘r’ is the common ratio.