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I'm getting stuck with a type of exercise on arithmetic progressions I never done before.

$\{u_n\}$ is an arithmetic progression:

  • $u_1+ u_2 + u_3 = 9$

  • $u_{10}+ u_{11} = 40$

I have then to prove that $u_0$ and the common difference $r$ are like:

  • $u_0+2r = 3$

  • $2u_0 + 21r = 40$

Finally, I can calculate $u_0$ and $r$.

I first thought to calculate $u_0$ and $r$ first and then trying to prove the equalities above. But that's not what it is asked in the exercise. I don't know how to manage this exercise.

What shall I do?

Thanks for your answers

mathlove
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Dhazard
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1 Answers1

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Since $u_1=u_0+r,u_2=u_0+2r,u_3=u_0+3r$, having $u_1+u_2+u_3=9$ gives you $$(u_0+r)+(u_0+2r)+(u_0+3r)=9\Rightarrow 3u_0+6r=9\Rightarrow u_0+2r=3.$$Also, since $u_{10}=u_0+10r,u_{11}=u_0+11r$, having $u_{10}+u_{11}=40$ gives you $$(u_0+10r)+(u_0+11r)=40\Rightarrow 2u_0+21r=40.$$

mathlove
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