Question

How do you find the first three terms of the arithmetic series $n = 19$, ${a_n} = 103$ and ${S_n} = 1102$.

Answer 1

We can solve this using the formula of sum of all terms in a finite arithmetic progression with the last term given. That is ${S_n} = \dfrac{n}{2}(a + l)$. Using thus we find the first term. Then we find the common difference using the ${n^{th}}$ formula that is ${a_n} = a + (n - 1)d$. Then we know that the general arithmetic progression form, $a,a + d,a + 2d,a + 3d...........,a + (n - 1)d$.

Complete step by step solution:
Given,
$n = 19$. Then we have
${a_{19}} = 103$ and ${S_{19}} = 1102$
We have a sum formula that is ${S_n} = \dfrac{n}{2}(a + l)$. Where ‘a’ is first term, ‘l’ is the last term and ‘n’ is the position of the term.
Put $n = 19$ in ${S_n} = \dfrac{n}{2}(a + l)$
${S_{19}} = \dfrac{{19}}{2}(a + l)$
Here last term is ${a_{19}} = 103(l)$ and substituting the given values we have,
$1102 = \dfrac{{19}}{2}(a + 103)$
Multiply by ‘2’ on both side we have,
$1102 \times 2 = 19(a + 103)$
$1102 \times 2 = 19a + \left( {19 \times 103} \right)$
$2204 = 19a + 1957$
$- 19a = - 2204 + 1957$
$- 19a = - 247$
Divide by -19 on both side we have,
$a = \dfrac{{ - 247}}{{ - 19}}$
$\Rightarrow a = 13$.
That is the first term is 13.
Now to find the common difference we have ${a_n} = a + (n - 1)d$. Where ‘a’ is the first term, ‘d’ is the common difference and ‘n’ is the position of the term.
Put $n = 19$ in ${a_n} = a + (n - 1)d$
${a_{19}} = a + (19 - 1)d$
${a_{19}} = a + 18d$
Substituting the values we have
$103 = 13 + 18d$
$103 - 13 = 18d$
Simplifying and rearranging we have
$18d = 90$
Dividing by 18 on both sides,
$d = \dfrac{{90}}{{18}}$
$\Rightarrow d = 5$.
Thus the common difference is ‘5’.
Now from the general form of A.P is $a,a + d,a + 2d,a + 3d...........,a + (n - 1)d$.
Thus the first three terms are $a,a + d$ and $a + 2d$
$a = 13$
$a + d = 13 + 5 = 18$
$a + 2d = 13 + 2(5) = 13 + 10 = 23$.
Hence the first three terms are 13,18 and 23.

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