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.

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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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. – Clarence Chew · 4 years ago

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– Aditya Parson · 4 years ago

I will corroborate your solution there.Log in to reply

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. – Ang Yan Sheng · 4 years ago

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there's a easy method using determinant criteria, for 3 points prove that the determinant of area equal to 0 – Pratik Singhal · 4 years ago

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– Aditya Parson · 4 years ago

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

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'. – Piyal De · 4 years ago

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– Aditya Parson · 4 years ago

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 \(x\).Log in to reply

– Piyal De · 4 years ago

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

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– Piyal De · 4 years ago

Buddy, onto means that f(x) will take all values in its range for x>=0, right? You showed that deno can never be = to zero for x>2/x<-2, so how can we say that it's onto from that..........as for equality, check the graph at x=-1, they're =. Actually the original prob was to find the range of f(x)=1/(x+2cos(x)), but I was sort of trying to solve it in another way so i asked the question.Log in to reply

– Aditya Parson · 4 years ago

then I don't think it can be onto.Log in to reply

– Aditya Parson · 4 years ago

\( cos 1\) is 0.540, not exactly 0.5.Log in to reply

– Piyal De · 4 years ago

Well actually I didn't do it..........but check the graph on a graphing cal.you'll see it as clear as crystal. Wat abt onto??Log in to reply

– Aditya Parson · 4 years ago

the co-domain is real numbers?Log in to reply

– Raja Metronetizen · 4 years ago

hey,you gave isi bmath exam,right...?Log in to reply

– Piyal De · 4 years ago

Nope friend, B.Stat............what about you?Log in to reply