Arithmetic Progression

**Tangera** · April 14th 2008, 07:03 AM

Hello! I just learnt arithmetic progression and I am confused about all the formula, so I would appreciate if someone showed me an example using the question below...Thank you very much!!

Q: In an arithmetic progression, the sum of the first 4 terms is 22 and the sum of the first 5 terms is 35. Find the hundredth term.

Thank you!

**angel.white** · April 14th 2008, 07:44 AM

Originally Posted by Tangera

Hello! I just learnt arithmetic progression and I am confused about all the formula, so I would appreciate if someone showed me an example using the question below...Thank you very much!!

Q: In an arithmetic progression, the sum of the first 4 terms is 22 and the sum of the first 5 terms is 35. Find the hundredth term.

Thank you!

$a_1 ~~=~~ a_1$

$a_2 ~~=~~ a_1 + x$

$a_3 ~~=~~ a_2 + x ~~=~~ a_1 +2x$

$a_4 ~~=~~ a_3 + x ~~=~~ a_1 +3x$

$a_5 ~~=~~ a_4 + x ~~=~~ a_1 +4x$

You were told the sum of the first 4 terms is 22, so:
$a_1 + (a_1+x) + (a_1+2x) + (a_1 +3x) ~~~~=~~~~ 4*a_1 +6x ~~~~=~~~~ 22$

You were told the sum of the first 5 terms is 35, so you can solve for the 5th term:
35-22 = 13, so $~~~~a_5 ~~~~=~~~~ a_1+4x ~~~~=~~~~ 13$

Now you have two equations and two variables.

$4*a_1 + 6x = 22$
$a_1 + 4x = 13~~~~~\Rightarrow a_1 = 13 - 4x$

Substitute this into the first equation
$4*(13-4x) + 6x = 22$

$52-10x = 22$

$x = 3$

And substitute this back into either equation
$a_1 + 4(3) = 13$
$a_1 = 1$

So now you have your values:
$a_1 = 1$
$a_2 = 1+3$
$a_3 = 1+3*2$
$a_4 = 1+3*3$
$a_5 = 1+3*4$

The pattern should be rather obvious: $a_n ~~~~=~~~~ 1+3(n-1) ~~~~=~~~~ 3n-2$

So $a_{100} ~~~~=~~~~ 3*100-2~~~~=~~~~ 298$

**Soroban** · April 14th 2008, 09:11 AM

Hello, Tangera!

In an arithmetic progression, the sum of the first 4 terms is 22
and the sum of the first 5 terms is 35. .Find the hundredth term.

We are expected to know these formulas . . .

$\text{Given the first term, }a\text{, and the common difference, }d$ :

. . $\text{The }n^{th}\text{ term is: }\;a_n \;=\;a + (n-1)d$

. . $\text{The sum of the first }n\text{ terms: }\;n \:=\:\frac{n}{2}[2a + (n-1)d]$

We are told: . $S_4\,=\,22$
. . So we have: . $\frac{4}{2}[2a + 3d] \:=\:22 \quad\Rightarrow\quad 2a + 3d \:=\:11\;\;{\color{blue}[1]}$

We are told: . $S_5\,=\,35$
. . So we have: . $\frac{5}{2}[2a + 4d] \:=\:35 \quad\Rightarrow\quad 2a + 4d \:=\:14\;\;{\color{blue}[2]}$

Subtract [1] from [2]: . $\boxed{d \:=\:3}$

Substitute into [1]: . $2a + 3(3) \:=\:11\quad\Rightarrow\quad\boxed{ a \:=\:1}$

Therefore: . $a_{100} \;=\;a + 99d \;=\;1 + 99(3) \quad\Rightarrow\quad\boxed{ a_{100}\;=\;298}$

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