# How do you find the first three terms of the arithmetic series a_1=17, a_n=197, and S_n=2247?

Oct 28, 2017

First three terms are $17 , 26 \mathmr{and} 35$

#### Explanation:

$1$st term is ${a}_{1} = 17 , {a}_{n} = 197$ , common difference is $b$

And number of terms is $n$.

Mid term is ${a}_{m} = \frac{17 + 197}{2} = 107$

Sum ${S}_{n} + = {a}_{m} \cdot n \mathmr{and} 2247 = 107 \cdot n \therefore n = \frac{2247}{107} = 21$

${a}_{n} = {a}_{1} + \left(n - 1\right) b \mathmr{and} 197 = 17 + \left(21 - 1\right) b \mathmr{and} 20 b = 180$ or

$b = \frac{180}{20} = 9 \therefore {a}_{2} = {a}_{1} + b = 17 + 9 = 26$

${a}_{3} = {a}_{1} + 2 b = 17 + 2 \cdot 9 = 17 + 18 = 35$

First three terms are $17 , 26 \mathmr{and} 35$ [Ans]

Oct 28, 2017

$17 , 26 , 35. . .$

#### Explanation:

An arithmetic sequence is the one in which the difference of the successive terms is the same

For example,

$1 , 3 , 5 , 7. .$ is an arithmetic sequence because the common difference is $2$. The symbol that i will use for the common difference here is $d$

Now lets solve the problem. For that we use the formula

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

Where, ${S}_{n}$ is the sum of the first $n$ terms, ${a}_{1}$ is the first term and ${a}_{n}$ is the ${n}^{t h}$ term. We already knew the values, so apply them

$\rightarrow 2247 = n \left(\frac{197 + 17}{2}\right)$

$\rightarrow 2247 = n \left(214\right)$

color(green)(rArrn=21

Now we know that the ${21}^{s t}$ term is $197$

To find the common difference, we use

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

$\rightarrow 197 = 17 + \left(21 - 1\right) d$

$\rightarrow 180 = 20 d$

color(green)(rArrd=9

Now we have come to an end to the answer. We add $9$ to each of the successive terms and we get

color(purple)(17,26,35....

Hope this helps!!! ☺