In an arithmetic progression, the sum of the first ten terms is 50, and the fith term is three times the second term. Find:
i) The first term
ii) The sum of the first 20 terms
Im having problems with finding the first term, I will be able to do the second part of the question easily, as its just the formula.

My text book hasnt got any examples of this sort of question in it, so Im unsure how to go about trying to complete it.

thanks

2. Hello fishkeeper
Originally Posted by fishkeeper
Im having problems with finding the first term, I will be able to do the second part of the question easily, as its just the formula.

My text book hasnt got any examples of this sort of question in it, so Im unsure how to go about trying to complete it.

thanks
With the usual notation, the sum of the first $10$ terms is:
$5(2a + 9d) = 50$

$\Rightarrow 2a+9d = 10$
...(1)
The second term is $a + d$; the fifth is $a + 4d$. So:
$a+4d=3(a+d)$

$\Rightarrow 2a = d$
Substitute into (1):
$10d=10$

$\Rightarrow d = 1$

$\Rightarrow a = \tfrac12$
OK now?

3. thanyou very much for a quick reply

Its taken me a little while to work through what you wrote, but I think ive understood fairly well, I'll write out my interpretation on the second part.

the $a+ 4d$: the $4d$ is the 4 terms that have each had the common ratio added onto them? And that equals 3 lots of $a+d$(the second term of the sequence as its had the common ratio added onto it?)

thanks

4. Hello fishkeeper
Originally Posted by fishkeeper
thanyou very much for a quick reply

Its taken me a little while to work through what you wrote, but I think ive understood fairly well, I'll write out my interpretation on the second part.

the $a+ 4d$: the $4d$ is the 4 terms that have each had the common ratio added onto them? And that equals 3 lots of $a+d$(the second term of the sequence as its had the common ratio added onto it?)

thanks
That's correct (although it's the common difference in an AP).