Question

Mid – point of A (0, 0) and B (1024, 2048) is \[{{A}_{1}}\], mid – point of \[{{A}_{1}}\] and B is \[{{A}_{2}}\]and so on. Co – ordinates of \[{{A}_{10}}\] are: -
(a) (1022, 2044)
(b) (1025, 2050)
(c) (1025, 2046)
(d) (1, 2)

Answer 1

Hint: Use the mid – point formula given by: -
Co – ordinates of mid – point = \[\left( \dfrac{{{x}_{1}}+{{x}_{2}}}{2},\dfrac{{{y}_{1}}+{{y}_{2}}}{2} \right)\], where \[{{x}_{1}}\] and \[{{x}_{2}}\] are the co – ordinates of A and B respectively and \[{{y}_{1}}\] and \[{{y}_{2}}\] are the y – coordinates of A and B respectively, to determine the coordinate of \[{{A}_{1}}\]. Similarly, use the formula to determine the coordinates of \[{{A}_{2}},{{A}_{3}}\] and so on. Form a geometric progression of both x and y – coordinates and use the formula: -
\[{{S}_{n}}=\dfrac{a\left( 1-{{r}^{n}} \right)}{\left( 1-r \right)}\], where ‘\[{{S}_{n}}\]’ is the sum of n terms of G.P, ‘a’ is the first term and ‘r’ is the common ratio. Substitute value of ‘n’ equal to 10.

Complete step by step answer:
We have been given two points A (0, 0) and B (1024, 2048). Using the mid – point formula given by: -
Mid – point = \[\left( \dfrac{{{x}_{1}}+{{x}_{2}}}{2},\dfrac{{{y}_{1}}+{{y}_{2}}}{2} \right)\], where \[{{x}_{1}}\] and \[{{x}_{2}}\] are the co – ordinates of A and B respectively and \[{{y}_{1}}\] and \[{{y}_{2}}\] are the y – coordinates of A and B respectively, we get,
x – coordinate of \[{{A}_{1}}\] = \[\dfrac{0+1024}{2}=\dfrac{1024}{2}\].
Now, \[{{A}_{2}}\] is the mid – point of \[{{A}_{1}}\] and B, therefore, x – coordinate of \[{{A}_{2}}\],
\[\begin{align}
  & =\dfrac{\dfrac{1024}{2}+1024}{2} \\
 & =\dfrac{1024}{2}+\dfrac{1024}{{{2}^{2}}} \\
\end{align}\]
On observing the series, we get that x – coordinate of \[{{A}_{3}}\] will be \[\dfrac{1024}{{{2}^{1}}}+\dfrac{1024}{{{2}^{2}}}+\dfrac{1024}{{{2}^{3}}}\] and so on.
Hence, x – coordinate of \[{{A}_{10}}\] will be,
\[\begin{align}
  & =\dfrac{1024}{{{2}^{1}}}+\dfrac{1024}{{{2}^{2}}}+\dfrac{1024}{{{2}^{3}}}+......+\dfrac{1024}{{{2}^{10}}} \\
 & =1024\left[ \dfrac{1}{{{2}^{1}}}+\dfrac{1}{{{2}^{2}}}+\dfrac{1}{{{2}^{3}}}+......+\dfrac{1}{{{2}^{10}}} \right] \\
 & ={{2}^{10}}\left[ \dfrac{1}{{{2}^{1}}}+\dfrac{1}{{{2}^{2}}}+\dfrac{1}{{{2}^{3}}}+......+\dfrac{1}{{{2}^{10}}} \right] \\
\end{align}\]
Clearly we can see that the terms inside the square brackets are in G.P with first term (a) equal to \[\dfrac{1}{2}\] and common ratio (r) = \[\dfrac{\dfrac{1}{{{2}^{2}}}}{\dfrac{1}{{{2}^{1}}}}=\dfrac{1}{2}\].
Now, using the formula of sum of ‘n’ terms of G.P, \[{{S}_{n}}=\dfrac{a\left( 1-{{r}^{n}} \right)}{\left( 1-r \right)}\], we get,
\[\begin{align}
  & \Rightarrow {{S}_{10}}=\dfrac{1}{2}\left( \dfrac{1-\dfrac{1}{{{2}^{10}}}}{1-\dfrac{1}{2}} \right) \\
 & \Rightarrow {{S}_{10}}=\left( 1-\dfrac{1}{{{2}^{10}}} \right) \\
\end{align}\]
Therefore, x – coordinate of \[{{A}_{10}}\] will be,
\[={{2}^{10}}\left( 1-\dfrac{1}{{{2}^{10}}} \right)\]
\[={{2}^{10}}-1\]
\[\begin{align}
  & =1024-1 \\
 & =1023 \\
\end{align}\]
Similarly, forming series of geometric progression for y – coordinate of \[{{A}_{10}}\], we have,
Coordinate of \[{{A}_{10}}\], we have,
\[\begin{align}
  & =\dfrac{2048}{{{2}^{1}}}+\dfrac{2048}{{{2}^{2}}}+.....+\dfrac{2048}{{{2}^{10}}} \\
 & =2048\left[ \dfrac{1}{{{2}^{1}}}+\dfrac{1}{{{2}^{2}}}+......+\dfrac{1}{{{2}^{10}}} \right] \\
\end{align}\]
Here also we can see that the terms inside the square brackets are in G.P whose sum we have calculate equal to \[\left( 1-\dfrac{1}{{{2}^{10}}} \right)\].
Therefore, y – coordinate of \[{{A}_{10}}\],
\[\begin{align}
  & =2048\left( 1-\dfrac{1}{{{2}^{10}}} \right) \\
 & =2048\left( 1-\dfrac{1}{1024} \right) \\
 & =2048-2 \\
 & =2046 \\
\end{align}\]
Therefore, the coordinates of \[{{A}_{10}}\] are (1023, 2046).

So, the correct answer is “Option C”.

Note: One may note that we have been provided with four options here. These options have different x and y – coordinates. So, if we are able to find any of x or y – coordinate then we will get the answer. This will reduce our time to solve this question. But be careful if there are no options or any coordinate of two option matches.