Question

how many terms of the arithmetic progression 9,17,25,33... must be taken to get a sum of 636

Answer 1

You need 12 terms to get a sum of 636 in the arithmetic progression.

1. This is the arithmetic progression:

9       17        25       33 ...          $x_{n}$

  +8        +8        +8       +8

2. There is a rule to get de last term of the progression:

$x_{n}=x_{1}+(n-1)d$

where:

$x_{n}$ Is the last term

$x_{1}$ Is de first term

$n$ Is the amount of terms

$d$ Is the difference between terms

We replace the values:

$x_{n}=9+(n-1)8\\\\x_{n}=9+8n-8\\\\x_{n}=8n+1$ ...(1)

3. The rule to get the SUM is:

$S=n(\frac{x_{1}+x_{n}}{1})$ .. (2)

We can replace (1) in (2)

$636=n(\frac{n(9+8n+1)}{2})\\\\636=n(\frac{10+8n}{2})\\\\636=5n+4n^2$

$5n+4n^2-636=0$

This kind of equation have this rule:

$ax^2+bx+c=0\\\\x=\frac{-b+\sqrt{b^2-4ac}}{2.4}$

Replacing values:

$n=\frac{-5+\sqrt{5^2-4.4(-636)}}{2.4}\\\\n=\frac{-5+\sqrt{25+10176)}}{8}\\\\n=\frac{-5+\sqrt{10201)}}{8}$

We have two options:

$n_{1}=\frac{-5+101}{8}=\frac{96}{8}=12\\\\n_{2}=\frac{-5-101}{8}=\frac{-106}{8}=-\frac{53}{4}$

We take the value n=12