Let \(a_{1}, a_{2}, a_{3},....a_{n}\) and \(b_{1} , b_{2}, b_{3},...b_{n}\) be two arithmetic progressions.

Then prove that the points : \( (a_{1},b_{1} ),(a_{2},b_{2}),.......,(a_{n},b_{n}) \) are collinear.

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## Comments

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TopNewestWe will now show that any 3 points are collinear.

Thus, we have to show \((a_k, b_k)\), \((a_m, b_m)\) and \((a_n, b_n)\) are collinear.

\(\iff\) Line formed by \((a_k, b_k)\) and \((a_m, b_m)\) is parallel to line formed by \((a_m, b_m)\) and \((a_n, b_n)\)

Using the fact of arithmetic progressions, we can let \(a_k=a_0+k \times d_a\) and \(b_k=b_0+k \times d_b\), for some constants \(a_0, d_a, b_0, d_b\)

Therefore we can show that the gradients of the 2 lines are equal.

Gradient of line formed by \((a_k, b_k)\) and \((a_m, b_m)=\frac{b_m-b_k}{a_m-a_k}=\frac{d_b}{d_a}\)

Gradient of line formed by \((a_m, b_m)\) and \((a_n, b_n)=\frac{b_m-b_n}{a_m-a_n}=\frac{d_b}{d_a}\)

Since gradients are equal, it follows that the lines are parallel and thus \((a_k, b_k)\), \((a_m, b_m)\) and \((a_n, b_n)\) are collinear.

Now we can start with base case n=3 and induct by taking points \((a_1, b_1), (a_2, b_2)\) and \((a_{n+1}, b_{n+1})\). Hence proven.

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I will corroborate your solution there.

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Let \(a_i = \lambda_a i+\mu_a\), \(b_i = \lambda_b i+\mu_b\) for some real numbers \(\lambda_a,\lambda_b,\mu_a,\mu_b\).

Then \(i=\frac{a_i-\mu_a}{\lambda_a}\), and thus \(b_i = \lambda_b\frac{a_i-\mu_a}{\lambda_a}+\mu_b = \frac{\lambda_b}{\lambda_a}a_i+(\mu_b-\frac{\mu_a\lambda_b}{\lambda_a})\).

Hence all the points \((a_i,b_i)\) lie on the same line \(y=\frac{\lambda_b}{\lambda_a}x+(\mu_b-\frac{\mu_a\lambda_b}{\lambda_a})\), as desired.

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there's a easy method using determinant criteria, for 3 points prove that the determinant of area equal to 0

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This method is far more convenient in exams, since it would take much less time. Determinant method is always applicable to show collinearity.

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This prob came in C.M.I....................... Do you know how to prove that f(x)=1/(x+2cos(x)) is onto rigorously...I mean intuitively it can be seen, but is there any othere way to do it.............I mean using 'hardcore math'.

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There exists \(f(x)\) whenever \(x+2cosx \neq=0\). But I don't know how can we conclude that \(x\) will never be \(=-2cosx\) for all \(x\).

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at x=-1........-x/2 and cos(x) are same.............so Domain,I guess, is R-{-1}......

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Comment deleted May 21, 2013

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hey,you gave isi bmath exam,right...?

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Nope friend, B.Stat............what about you?

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