How do you find the first three terms of the arithmetic series n=19, a_n=103, S_n=1102?

May 3, 2018

See explanation.

Explanation:

First we have to write everything which is given and what we are looking for:

Given:

$n = 19$

${a}_{19} = 103$

${S}_{19} = 1102$

To calculate:

${a}_{1}$, ${a}_{2}$, ${a}_{3}$

First we can use the sum formula to calculate ${a}_{1}$:

${a}_{1} = 13$

Now we can calculate the common difference using two given terms:

$d = 5$

Now having ${a}_{1}$ and $d$ we can calculate ${a}_{2}$ and ${a}_{3}$:

${a}_{2} = {a}_{1} + d = 13 + 5 = 18$

${a}_{3} = {a}_{2} + d = 18 + 5 = 23$

The first three terms are: $13 , 18$ and $23$.

May 3, 2018

$13 , 18 , 23.$

Explanation:

Here,

$n = 19 , {a}_{n} = 103 , {S}_{n} = 1102$.

We know that,

${n}^{t h} t e r m$ of Arithmetic series is$= {a}_{n}$

and sum of first n-terms$= {S}_{n}$

Now, $a =$first term and

$d =$common ratio of Arithmetic series.

So,

color(red)(a_n=a+(n-1)d and S_n=n/2[2a+(n-1)d]

$\implies 103 = a + \left(19 - 1\right) d \mathmr{and} 1102 = \frac{19}{2} \left[2 a + \left(19 - 1\right) d\right]$

$\implies 103 = a + 18 d \mathmr{and} 1102 \times \frac{2}{19} = 2 a + 18 d$

$\implies 103 = a + 18 d \mathmr{and} 116 = 2 a + 18 d$

Let,

color(blue)(a+18d=103 to(1)and 2a+18d=116to(2)

From $\left(2\right)$ we get,

$a + \textcolor{b l u e}{a + 18 d} = 116. . . \to \left[2 a = a + a\right]$

a+color(blue)(103)=116...tocolor(blue)([use (1) ]

$\implies a = 116 - 103$

=>color(violet)(a=13

From $\left(1\right)$we have,

$13 + 18 d = 103$

$\implies 18 d = 103 - 13$

$\implies 18 d = 90$

=>color(violet)(d=5

Hence, first three terms of Arithmetic series are :

(i)a=color(brown)(13

(ii)a+d=13+5=color(brown)(18

(iii)a+2d=13+2xx5=13+10=color(brown)(23

May 3, 2018

$\text{ }$
The first three terms: color(blue)(a_1 = 13, a_2=18 and a_3=23

Explanation:

$\text{ }$
Total number of terms: color(red)(n=19

19th term: color(red)(a_19=103

Sum of the first 19 terms: color(red)(S_19=1102

In an Arithmetic Sequence, the difference between one term and the next is a Common Difference: $\textcolor{red}{d}$.

The terms are:

color(brown)(a_1, (a_1+d), (a_1 + 2d), (a_1+3d), ... , where color(brown)(a_1 is the First Term and color(brown)(d is the Common Difference.

The Sum of an arithmetic sequence is called an Arithmetic Series.

color(green)("Step 1:"

color(blue)("Formula 1:"

Sum to $\textcolor{red}{n}$ terms of an arithmetic series:

color(green)(S_n = [n(a_1+a_n)]/2.

color(blue)("Formula 2:"

Find the color(red)(n^(th) term of the arithmetic series:

color(green)(a_n=a_1+d(n-1)

color(green)("Step 2:"

Since, $\textcolor{red}{n , {a}_{n} , \mathmr{and} {S}_{n}}$ are given, use color(blue)("Formula 1" to find $\textcolor{red}{{a}_{1}}$.

$1102 = \frac{19 \left({a}_{1} + {a}_{19}\right)}{2}$

$1102 = \frac{19 \left({a}_{1} + 103\right)}{2}$

$1102 = \frac{19 {a}_{1} + 1957}{2}$

Multiply both sides of the equation by color(red)(2.

1102*color(red)(2) =(19a_1+1957)/cancel 2*color(red)(cancel 2

$2204 = 19 {a}_{1} + 1957$

Flipping sides:

$19 {a}_{1} + 1957 = 2204$

Subtract color(red)(1957.

$19 {a}_{1} + \cancel{1957} - \textcolor{red}{\cancel{1957}} = 2204 - \textcolor{red}{1957}$

$19 {a}_{1} = 247$

Divide both sides by color(red)(19

(cancel 19a_1)/color(red)(cancel 19)=247/color(red)(19

${a}_{1} = \frac{247}{19} = 13$

$\therefore \text{First Term } = {a}_{1} = 13$

color(green)("Step 3:"

Use color(blue)("Formula 2" to find the Common Difference $\textcolor{red}{d}$.

color(green)(a_n=a_1+d(n-1)

When $n = 19 , {a}_{n} = {a}_{19}$

$103 = 13 + d \left(19 - 1\right)$

$103 = 13 + 19 d - d$

$103 = 13 + 18 d$

Flipping sides:

$13 + 18 d = 103$

Subtract $\textcolor{red}{13}$.

$\cancel{13} + 18 d - \textcolor{red}{\cancel{13}} = 103 - \textcolor{red}{13}$

$18 d = 90$

Divide by color(red)(18.

(cancel 18d)/color(red)(cancel 18)=cancel 90^color(green)(5)/color(red)(cancel 18

 :. "Common Difference " color(red)(d=5

Terms: $13 + \left[13 + 1 \left(5\right)\right] + \left[13 + 2 \left(5\right)\right]$

So, ${a}_{1} = 13 , {a}_{2} = 18 \mathmr{and} {a}_{3} = 23$