[Math] How to find first term, common difference, and sum of an arithmetic progression

algebra-precalculuscalculussequences-and-series

Given that an arithmetic progression is such that the 8th term is twice the second term, and the 11th term is 18.
Find:
1) The first term and common difference.
2) The sum of the first 26 terms.
3) The smallest of the progression whose values exceed 126?

How on earth am I meant to solve this? I'm guessing you try and find a formula for the nth term, but I have no clue how to get there.
Any suggestions?

Best Answer

First we write out our given information:

$$a_8=2a_2$$

$$a_{11}=18$$

$a_n$ is an arithmetic sequence.

Where here $a_n$ means the $n$th term of our sequence.

What does an arithmetic sequence mean? It means to get to the next term in your sequence you add a constant ($c$) each time:

$$a_{n+1}=a_n+c$$

Equivalently:

$$\frac{a_{n+1}-a_{n}}{(n+1)-n}=c$$

So $a_n$ is of slope $c$ ($c_2$ is another constant):

$$a_n=cn+c_2$$

Where here $c_2=a_0$ (Substitute in $n=0$ and see why that has to be the case if we let $a_0$ exist)

Now we use the other given information to try to come up with a solution.

Let $n=2$:

$$a_2=2c+c_2 {}{}$$

Let $n=8$, using the above equation we have:

$$a_8=8c+c_2=2a_2=4c+2c_2 {}{}{}{}$$

Let $n=11$

$$a_{11}=18$$

$$a_{11}=11c+c_2$$

But $a_{11}-a_8=(11c+c_2)-(8c+c_2)=3c$

Hence, $a_{11}=3c+a_{8}$

$$a_{11}=3c+4c+2c_2=18$$

$$a_{11}=3c+8c+c_2=18$$

Solve this system of equations to get a closed form for the arithmetic sequence.

$$a_n=1.2n+4.8$$.
We can check it works $a_2=1.2(2)+4.8=7.2$. Now we compute $a_8$ to see if $a_8=2a_2$ as required: $a_8=1.2(8)+4.8=14.4=2(7.2)=2a_2$. It is arithmetic as we may check $a_{n+1}-a_n$ is a constant $1.2$. Also $a_{11}=1.2(11)+4.8=18$ as required.

The answers follow from this, from summation formulas/methods of evaluating sums, and from algebra.

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