Question

the alegebraic expression of an arithmetic sequence is sn-3
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Answer 1

Answer:

1=1a1=1

2=7=1+1.6a2=7=1+1.6

3=13=1+2.6a3=13=1+2.6

\displaystyle a_n=1+(n-1).6an=1+(n−1).6 - general term of arithmetic progression with difference \displaystyle d=6d=6

\displaystyle a_1+a_2+a_3+\cdots+a_n=S_n=\frac{2a_1+(n-1).d}{2}\cdot n=\frac{2+(n-1).6}{2}\cdot n=(1+3(n-1)).n=(3n-2).na1+a2+a3+⋯+an=Sn=22a1+(n−1).d⋅n=22+(n−1).6⋅n=(1+3(n−1)).n=(3n−2).n

\displaystyle S_n=833Sn=833

\displaystyle n(3n-2)=833n(3n−2)=833

\displaystyle 3n^2-2n-833=03n2−2n−833=0 - quadratic equation for n with abbreviated discriminant \displaystyle D_1=\left(\frac{-2}{2}\right)^2-3.(-833)=1+2499=2500=50^2D1=(2−2)2−3.(−833)=1+2499=2500=502

\displaystyle n=\frac{-\frac{b}{2}\pm\sqrt{D_1}}{a}=\frac{-\frac{-2}{2}\pm50}{3}=\frac{51}{3};\ -\frac{49}{3}n=a−2b±D1=3−2−2±50=351; −349

\displaystyle n>0\Rightarrow n=\frac{51}{3}=17n>0⇒n=351=17

Explanation:

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