Find the 20th member of an arithmetic progression

Question

The progression is increasing arithmetic

I am given that $a_2+a_3+a_4+a_5=34$ and that $a_2\cdot a_5=52$ I find that $a_5=13$ or $4$ it can't be $4$ so when i try with $a_5=13$ the $a_{20}=43$ and the answer is $58$

Answer 1

If you write down the system of equations corresponding to the constraints you provide you have ($h$ is the size of the progression):

$ \left\{ \begin{align*} 4\cdot a_2+ 6h &= 34\\ a_2(a_2+3h)&=52 \end{align*} \right. $

You find the following two solutions

$ \begin{align*} a_2 &= 4\\ h &=3 \end{align*} $

and

$ \begin{align*} a_2 &= 13\\ h &=-3 \end{align*} $

Assuming $h=3$ one finds

$a_{20}=a_2+18h=4+54=58$.

Answer 2

$a_k=a_0+kb$, so:

$\begin{cases}a_2+a_3+a_4+a_5 = 4a_0+14 b = 34\\ a_2\cdot a_5 = (a_0 + 2b)(a_0 + 5b)=52\end{cases}$

From the first equation we obtain $a_0=\frac{17-7b}{2}$, so

$a_0^2 + 7 a_0 b + 10b^2-13=0$

is equivalent to

$\left(\frac{17-7b}{2}\right)^2+ 7b\frac{17-7b}{2}+10b^2-52=0\\ 289- 238b + 49b^2 + 238 b -98b^2+40b^2 - 208 =0 \\ -9b^2 + 81 =0 \\ b^2-9=0 $

Thus:

$\begin{cases}b= \pm 3\\ a_0 = \frac{17\mp 21}{2} \end{cases}$

and

$a_{20}=\frac{17 \mp 21}{2} \pm 20\cdot 3 = \frac{17 \pm 99}{2}$

Thus $a_{20}=58$ or $a_{20}=-41$