EDIT –
here's a screenshot of the question
I'm trying to do a series problem and keep on going wrong. The question is :
$$\text{The sum of the first two terms of an arithmetic series is 47}$$
$$\text{The 13th term of this series is -62. Find:}$$
$$\text{a. the first term of the series and the common difference }$$
$$\text{b. the sum of the first 60 terms of the series}$$
So what I have tried is as follows :
The question has given the sum of the first two terms, so I should be able to use this as a series in itself, and say that :
$$a = a$$
$$n = 2$$
$$d = d$$
$$sum = 47$$
$$47 = 2a + d $$
Then looking at the other piece of information the 13th term is -62. So I know that it's going to be a negative common difference, and that -62 will be :
$$-62 = a + (n-1)d$$
$$-62 = a + 12d$$
I can then use simultaneous equations to solve for a and d.
$$\text{eq[1]: 47 = 2a + d} $$
$$\text{eq[2]: -62 = a + 12d} $$
$$\text{multiply equation 2 by 2 then calculate eq[1] – eq[2]}$$
$$\text{equation 3 : }2(eq[2]) = -124 = 2a + 24d$$
$$\text{then work out eq[1] – eq[3] which leaves :}$$
$$171 = 0a – 23d$$
$$\text{So I should be able to say that the value for d is }-\frac{171}{23}$$
But this answer is wrong, it also looks wrong. But that's just because it's not a nice number and the questions in my level usually have nice number answers, which probably isn't great logic going forwards but still.
I'm not sure what I'm doing wrong here, no doubt something very basic.
Thank you
Best Answer
Actually the question is saying that the $30th$ element of the progression is $-62$.
For the record, I find ordinal numbers' names horribly confusing. In fact, I didn't even know that a thing such as thirtieth existed!
Taking this into account, we have the following system $$\begin{cases}47 = 2a + d\\ -62 = a + 29d \end{cases}$$ which indeed gives $a = 25$ and $d = -3$.
But remember: beyond the actual problem the most important thing is the method, and yours was working (albeit for a different version of the problem).