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This question is part of my advanced maths assignment on arithmetic and geometric progressions.

Here's the question:
Q4. Given that both the sum of the first ten terms of an arithmetic progression and the sum of the 11th and 12th terms of the same progression are equal to 60, find the first term and the common difference.
Here's what I've done so far:
Working out for question 4
I've done this question about four times and I'm still having the same result.

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    $\begingroup$ Should be $20a+90d=120$, not $20a+45d=120$, I believe you forgot to multiply $45d$ by 2 $\endgroup$
    – Itakura
    Apr 30 '16 at 12:50
  • $\begingroup$ You could avoid some of your calculations by dividing $60 = 5(2a + 9d)$ by $5$ to obtain $12 = 2a + 9d$. Then you have to solve the system of equations \begin{alignat*}{3} 2a & + & 9d & = 12\\ 2a & + & 21d & = 60 \end{alignat*} which is easy to solve for $d$. Fewer calculations mean fewer opportunities to make an error. $\endgroup$ Apr 30 '16 at 13:27
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Here is what I got:

For the first 10 terms and $S_{10}=60$, we have $S_{10}=\frac{10}{2}\left(2u_1+9d)\right)=10u_1+45d$

For the 11th term and 12th term we have $u_{11}+u_{12}=60$ as well. So we have:

$u_{11}=u_1+10d$ and $u_{12}=u_1+11d$ and substituting these values will give:

$60=2u_1+21d$

Now since $10u_1+45d=60$ and $2u_1+21d=60$ as well, we can end up with this equation:

$$10u_1+45d=2u_1+21d$$

$$8u_1=-24d$$

$$u_1=-3d$$

Substituting $u_1=-3d$ to either equations $10u_1+45d=60$ or $2u_1+21d=60$ will still yield same result.

$$2(-3d)+21d=60$$

$$21d-6d=60$$

$$15d=60$$

$$d=4$$

So we have $u_1=-3(4)=-12$

So conclusion:$u_1=-12$ and $d=4$

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