Check if $6, 6-\sqrt{2}, 6-2\sqrt{2}, 6-3\sqrt{2}$ forms an Arithmetic progression.If it does then write the common difference and the next two terms. I know the answer but please tell me the proof and explain it.
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closed as not a real question by Andres Caicedo, tomasz, Arkamis, Norbert, Noah Snyder Oct 13 '12 at 21:47
It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, see the FAQ.
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I’ll do a slightly different problem as a model. Suppose instead that the sequence was $$20,20-\pi,20-2\pi,20-3\pi\;.$$ Then $$\begin{align*} &(20-\pi)-20=-\pi\;,\\ &(20-2\pi)-(20-\pi)=-\pi\;,\text{ and}\\ &(20-3\pi)-(20-2\pi)=-\pi\;. \end{align*}$$ The differences of consecutive terms are all the same: each is $-\pi$, so this is an arithmetic progression with common difference $-\pi$. To get the next term, we just add the common difference to the last term that we have: $$20-3\pi+(-\pi)=20-4\pi\;.$$ Doing it again gives us the sixth term: $$20-4\pi+(-\pi)=20-5\pi\;.$$ |
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As you go from $6$ to $6 - \sqrt{2}$, what do you add (it could be negative)? As you go from $6 - \sqrt{2}$ to $6 - 2\sqrt{2}$, did you add the exact same thing? And, finally, as you go from $6 - 2\sqrt{2}$ to $6 - 3\sqrt{2}$, was it again the same? If so, it is an arithmetic progression, and this amount you added each time is the common difference. And, to find the next term, just add that same common different to the last term you currently have. |
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It's actually a pretty big step in a novice mathematician's life when he/she gains an intuitive notion for what qualifies as a proof... so I wouldn't assume that you're missing the idea and more that you don't understand what qualifies as a proof. However, in this case, the question doesn't ask for a proof. It says "if it is" then give him the common difference and next two terms. Those are requested without proof. If you want to "prove it's an arithmetic progression" then you need to do a few more steps. Foremost in most proofs are definitions. In this case you want the definition of an arithmetic progression. Then you need to "prove it's an arithmetic progression" by showing it satisfies that definition. That's all you need to do in this case. The general/intuitive/non-rigorous idea behind proofs is that you need a logical flow from known ideas to the final claim. |
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Write this: It is an arithmetic progression. The starting member $a_1=6$, and the difference is $d=-\sqrt2$. $$a_2=6-\sqrt2=a_1+d, \quad a_3=a_1+2d,\ ..., \quad a_n = a_1+(n-1)d$$ So, $a_5=6-4\sqrt2$ and $a_6=6-5\sqrt2$. |
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Many expositions on Arithmetic Progressions (e.g. Wikipedia) fail to emphasize their geometric essence. They are simply discrete analogs of lines (or line segments). In particular, the sequence $\rm\:y_0,y_1,y_2,\:$ is an arithmetic progression iff the points $\rm\:(0,y_0),\,(1,y_1),\,(2,y_2),\ldots\,\ldots$ are collinear. Now, from high-school, we know various techniques to check collinearity of points. For example, given the points $\rm\:(x_0,y_0),\,(x_1,y_1),\,(x_2,y_2),\ldots,\:$ with $\rm\:x_i < x_{\,i+1},\:$ it suffices to check constancy of the successive slopes $\rm\:(y_{i+1}\!-y_{i})/(x_{\,i+1}\!-x_i).\:$ In particular, in the AP case where $\rm\:x_{\,i+1}\!-x_{\,i} = 1,\:$ this amounts to checking constancy of the successive differences of $\rm y$ values, i.e. $\rm\:y_{i+1}\! - y_i = $ constant. |
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