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?

  • 2
    $\begingroup$ In an arithmetic sequence, where the first term is $a$, and where the common difference is $d$, the $n$-th term is $a+(n-1)d$. $\endgroup$
    – Kenny Lau
    Aug 17, 2016 at 3:47
  • $\begingroup$ Yep. Use what Kenny said to get a system of two equations in two unknowns ($a$ and $d$). $\endgroup$ Aug 17, 2016 at 3:48
  • $\begingroup$ 'Nother zombie thread. I fall for them every time. $\endgroup$
    – fleablood
    Nov 2, 2020 at 17:20

2 Answers 2


First we write out our given information:



$a_n$ 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:




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


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 $n=11$



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

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



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

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.


From $$\begin{cases}a+7d=2(a+d),\\a+10d=18\end{cases}$$



$$6+(n-1)\frac65>126\implies n\ge102.$$


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