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I was given this question on a college assessment pre-test. I got the correct answer in a reasonable amount of time, but mostly because I worked backwards and double checked my answer. After I was done, I tried to find math on the net to solve it correctly, but I feel there is probably a simpler solution than this.

Is there an easier way to solve problems in this template?

Given the first and n-th values in an arithmetic progression, and the sum of the progression up to n (inclusive), give the first x terms of the series.

The actual question on the quiz

In an arithmetic series, the terms of the series are equally spread out. For example, in 1 + 5 + 9 + 13 + 17, consecutive terms are 4 apart. If the first term in an arithmetic series is 3, the last term is 136, and the sum is 1,390, what are the first 3 terms?

This one lended itself to intuition and backwards-work because I was reasonably certain that d would be an integer, and 133 (136 - 3) is evenly divisible by 7. But I would like to find a simpler way to solve problems like this in the future.

The work I did to solve the question

I used the formulas:

  • Sn = ½ n(a1 + an)
  • an = dn + c

Where n is the count of values in the sequence, d is the common difference (distance between values in the sequence), and c is an unknown constant

For the first equation, I solved for n:

1390 = ½ n(3 + 136)
1390 = ½ n(139)
10 = ½ n
20 = n

For the second equation, I first solved for c, by making two equations out of the values I had for an, and the previously calculated value for n:

3 = d + c,
136 = 20d + c

(with the first): d = 3 - c
(substitute): 136 = 20(3 - c) + c
136 = 60 - 19c
76 = -19c
-4 = c

Then I solved for d:

(substitute): 3 = d - 4
d = 7
(double-check): 136 = 20d - 4
140 = 20d
7 = d

The final answer was 3, 3 + d, 3 + 2d:

3, 10, 17

Which I did a brute force double-check on.

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3  
Suppose you have $n$ terms, with the difference =$d$. Then $a_1+a_n=2a_1+(n-1)d$, and $\Sigma_{i=1}^n a_i=na_1+n(n-1)d/2$. Now solve it. For example, you know $(n-1)d=136-3=133$, and hence $1390=n(3+133/2)$, thus you have $n$, hence all the terms now. –  awllower Feb 10 '13 at 2:00

4 Answers 4

up vote 3 down vote accepted

Your approach is good. The only simplification that I can see is to get rid of that rather uninteresting variable $c=-4$ and compute directly in terms of $a_1=3$, by using $a_n = a_1 + (n-1)d$ instead of $a_n = c+nd$.

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That's exactly the intuitive answer I was looking for. Thanks! :) –  Merlyn Morgan-Graham Mar 17 '11 at 9:53

The average term is $\frac{3+136}{2}$, so there must be $\frac{1390}{139/2}=20$ terms in the sequence. What must the step size be, then? (Watch out for fencepost errors).

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The basic formulae for an arithmetic sequence with first element $\,a_1\,$ , common difference $\,d\,$ and the sum of the first $\,n\,$ consecutive elements $\,S_n\,$ are:

$$a_m=a_1+(m-1)d\;\;\;\;,\;\;S_m=\frac{m}{2}\left(2a_1+(m-1)d\right)=\frac{m}{2}(a_1+a_m)$$

In you case, we have $\,n\,$ terms , $\,a_1=3\,$ , and:

$$\begin{align*}I&\; a_n=136\\II&S_{36}=1,390\end{align*}$$

so using the above formulae:

$$\begin{align*}I\;\;&3+(n-1)d=136\\II\;\;&\frac{n}{2}(3+136)=1,390\end{align*}$$

thus

$$II\;\;\;\Longrightarrow\frac{n}{2}=\frac{1390}{139}=10\Longrightarrow n=20\Longrightarrow $$

$$I\;\;\;\Longrightarrow 3+19d=136\Longrightarrow d=\frac{133}{19}=7$$

And the first elements are $\,a_1\;\;,\;\;a_1+d\,\,,\,\,a_1+2d\ldots$

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If there is given number of elements $n$, first term $a_1$ and $n$-th term $a_n$, then sum of first $n$-th terms can be calculated as $S_n=\frac{n}{2}\left(a_n+a_1\right)$. From it you get $n=\frac{2S_n}{a_n+a_1}$.

When you have $n$, from $a_n=a_1+\left(n-1\right)d$ you get $d=\frac{a_n-a_1}{n-1}$.

And then it's simple - $a_2=a_1+d$ and $a_3=a_2+d$.
(If you want to check your calculations, the answer is $a_1=3$, $a_2=10$, $a_3=17$)

The main thing in such exercises, imho, is to get value of any term with known index $n$ and the difference $d$ - then you can calculate everything easily.

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