# In an arithmetic sequence, the third term is 10 and the fifth term is 16.

I'm just working on some summer problems so that I can be more prepared when I go into my class in the fall. I found a website full of problems of the content we will be learning but it doesn't have the answers. I need a little guidance on how to do this problem.

In an arithmetic sequence, the third term is $10$ and the fifth term is $16$.
Find the common difference.
Find the first term.
Find the sum of the first $20$ terms in the sequence.

So, the arithmetic formula is $a_n = a_1 + (n – 1)d$ right?
The common difference is the difference between the terms I think. So $16 - 10 = 6$, but there is a term between that so divided by $2$ it is $3$.

How do I find the first term and the sum of the first 20 terms?

• You've done well so far. If the common difference is $3$ and the third term is $10$, what is the second term? Once you answer that, you'll find the rest is easy. Jul 31, 2018 at 15:11
• @saulspatz Oh! So, the second term is 7 and the first term would be 4.
– Ella
Jul 31, 2018 at 15:15
• There you go, that's it! Jul 31, 2018 at 15:18

The pedestrian approach is to use your formula $a_n=a_1+(n-1)d$ and plug in the two values you know $$a_3=10=a_1+(3-1)d\\a_5=16=a_1+(5-1)d$$ This is two simultaneous equations that you can solve for $a_1,d$. You have already found $d=3$ essentially by subtracting the two. Now just plug that into one of them and evaluate $a_1$.

Otherwise, since you know $d=3$ you can just count two steps down from $a_3=10$ to get $a_1$

To get the sum of the first $20$ terms you need to know (and should have proved) that $$\sum_{i=1}^ni=\frac 12n(n+1)$$ If you write out the sum of the first $20$ terms you will need $d$ times this with $n=19$ plus to account for the $a$s

• You meant to write $$\sum_{i = 1}^{n} i = \frac{1}{2}n(n + 1)$$ Also, check your last sentence, which seems to suggest that multiplying this sum by $d$ will yield the sum of the first $20$ terms of the arithmetic sequence. Aug 1, 2018 at 17:19
• @N.F.Taussig: Thanks, fixed, and a remark about the $a$s added Aug 1, 2018 at 19:53
• I completed the whole problem and just wanted some clarification, if that was okay. For c, I got 650. Does this sound correct?
– Ella
Aug 4, 2018 at 2:44
• Yes, that is correct. Aug 4, 2018 at 4:06

You have correctly found the common difference.

The $k$th term of an arithmetic sequence with initial term $a_1$ and common difference $d$ is $$a_k = a_1 + (k - 1)d$$ Since we are given that $a_3 = 10$ and $a_5 = 16$, we obtain \begin{align*} a_3 & = a_1 + (3 - 1)d = 10\\ a_5 & = a_1 + (5 - 1)d = 16 \end{align*} which is a system of two linear equations in two variables. Subtracting the first equation from the second yields $2d = 6 \implies d = 3$, as you found. Substituting $d = 3$ in the equation for $a_3$ yields $a_1 = 4$, as you found in the comments.

The $n$th partial sum (the sum of the first $n$ terms) of an arithmetic series \begin{align*} S_n & = \sum_{k = 1}^{n} a_k\\ & = \sum_{k = 1}^{n} [a_1 + (k - 1)d]\\ & = \sum_{k = 1}^{n} a_1 + \sum_{k = 1}^{n} (k - 1)d\\ & = a_1\sum_{k = 1}^{n} 1 + d\sum_{k = 1}^{n} (k - 1)\\ & = na_1 + \frac{d(n - 1)n}{2} \end{align*} from which you can find $S_{20}$ by substituting $20$ for $n$, $3$ for $d$, and $4$ for $a_1$.

Alternatively, \begin{align*} S_n & = \sum_{k = 1}^{n} a_k\\ & = a_1 + a_2 + a_3 + \cdots + a_{n - 2} + a_{n - 1} + a_n\\ & = a_1 + [a_1 + d] + [a_1 + 2d] + \cdots + [a_1 + (n - 3)d] + [a_1 + (n - 2)d] + [a_1 + (n - 1)d] \end{align*} Since we obtain the same sum if we write the terms in reverse order, we obtain \begin{align*} S_n & = a_n + a_{n - 1} + a_{n - 2} + \cdots + a_3 + a_2 + a_1\\ & = [a_1 + (n - 1)d] + [a_1 + (n - 2)d] + [a_1 + (n - 3)d] + \cdots + [a_1 + 2d] + [a_1 + d] + a_1 \end{align*} Adding the two expressions for $S_n$ yields \begin{alignat*}{10} S_n & = & a_1 & + & [a_1 + d] & + & \cdots & + & [a_1 + (n - 1)d] & + & [a_1 + (n - 1)d]\\ S_n & = & [a_1 + (n - 1)d] & + & [a_1 + (n - 2)d] & + & \cdots & + & [a_1 + d] & + & a_1\\ \hline 2S_n & = & [2a_1 + (n - 1)d] & + & [2a_1 + (n - 1)d] & + & \cdots & + & [2a_1 + (n - 1)d] & + & [2a_1 + (n - 1)d] \end{alignat*} Since there are $n$ columns in that sum, we obtain \begin{align*} 2S_n & = n[2a_1 + (n - 1)d]\\ 2S_n & = n[a_1 + a_1 + (n - 1)d]\\ 2S_n & = n(a_1 + a_n)\\ S_n & = \frac{n(a_1 + a_n)}{2} \end{align*} from which you can find $S_{20}$ by substituting $4$ for $a_1$ and using the formula $a_n = a_1 + (n - 1)d$ with $n = 20$ and $d = 3$ to find $a_{20}$.