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The sum of the first $15$ terms of an arithmetic progression is $100$ and its $10$th term is $5$. Find the $5$th term and then calculate the sum of the first $50$ terms.

It is my understanding that in-order to find the nth term of an arithmetic progression, the formula is: $$T_n=a+(n-1)d$$ so $$T_{10}=a+9d=5$$ However, I do not know the first number nor do I know the difference.

What is the correct formula to solve this question?

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  • $\begingroup$ So what, in terms of $a$ and $d$, is the sum of the first fifteen terms? $\endgroup$ – Lord Shark the Unknown Mar 6 '18 at 17:52
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Guide:

$$\frac{15}{2}\left(2a+(15-1)d\right)=100$$

$$a+(10-1)d=5$$

Solve for $a$ and $d$ first, then you can solve your problem.

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you have the equations $$a_1+a_1+d+a_1+2d+...+a_1+14d=100$$ and $$a_1+9d=5$$ can you finish?

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The sum of the first $n$ terms of an arithmetic progression is

$$\frac{n(a_1+a_n)}{2}.$$ So we have

$$\frac{15(a_1+a_1+(14)d)}{2}=100,$$

and combining this with the fact that you have $$a_1+9d=5$$

gives you two equations with two unknowns...

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