Question

If \[a,b,c,d\] are in G.P., then show that, \[{(a - d)^2} = {(b - c)^2} + {(c - a)^2} + {(d - b)^2}\].

Answer 1

- Hint: A sequence of numbers is called a geometric progression if the ratio of any two consecutive terms is always the same.
The general form of G.P. is \[a,ar,a{r^2},...,a{r^n}\]. Where, \[a\]is the initial term and \[r\] be the common ratio.
First, we will represent the term of G.P. series in terms of the initial term and the common ratio.
Then we will try to prove the given problem.

Complete step-by-step solution -
It is given that, \[a,b,c,d\] are in G.P.
We have show that, \[{(a - d)^2} = {(b - c)^2} + {(c - a)^2} + {(d - b)^2}\]
First, we will represent the term of G.P. series in terms of the initial term and the common ratio.
Let us consider, \[p\] be the initial term and \[r\] be the common ratio of the given G.P. series.
So, we have,
\[a = p,b = pr,c = p{r^2},d = p{r^3}\]
Now, substitute these values into the right hand side of the given problem we have,
\[
  {(b - c)^2} + {(c - a)^2} + {(d - b)^2} \\
   = {(pr - p{r^2})^2} + {(p{r^2} - p)^2} + {(p{r^3} - pr)^2} \\
 \]
Now, we will apply the formula of \[{(a - b)^2} = {a^2} - 2ab + {b^2}\] we have,
\[ = {p^2}{r^2} - 2{p^2}{r^3} + {p^2}{r^4} + {p^2}{r^4} - 2{p^2}{r^2} + {p^2} + {p^2}{r^6} - 2{p^2}{r^4} + {p^2}{r^2}\]
Now we can eliminate the same term with the opposite sign.
We get,
\[ = - 2{p^2}{r^3} + {p^2} + {p^2}{r^6}\]
Simplifying we get,
\[ = {p^2}{r^6} - 2{p^2}{r^3} + {p^2}\]
This above expression can be written as,
\[ = {(p{r^3})^2} - 2p{r^3}.p + {p^2}\]
Again, applying the formula of \[{(a - b)^2} = {a^2} - 2ab + {b^2}\]we have,
\[ = {(p{r^3} - p)^2}\]
Substitute the values of \[a = p,d = p{r^3}\] we have,
\[ = {(d - a)^2}\]
Therefore, the given left hand side and right hand side are equal.
Hence,
\[{(a - d)^2} = {(b - c)^2} + {(c - a)^2} + {(d - b)^2}\] (Proved)

Note: This problem can be solved in another way.
Since, \[a,b,c,d\] are in G.P.
We can write the terms as,
\[\dfrac{a}{b} = \dfrac{c}{d}\]
We get, \[ad = bc\]
Also, \[ac = {b^2},bd = {c^2}\]
Now, Now, we will apply the formula of \[{(a - b)^2} = {a^2} - 2ab + {b^2}\] at the right hand side of the given problem, we have,
\[
  {(b - c)^2} + {(c - a)^2} + {(d - b)^2} \\
   = {b^2} - 2bc + {c^2} + {c^2} - 2ca + {a^2} + {d^2} - 2bd + {b^2} \\
 \]
Simplifying we get,
\[ = {a^2} + 2{b^2} + 2{c^2} + {d^2} - 2ad - 2ac - 2bd\]
Substitute the value \[ad = bc\] and \[ac = {b^2},bd = {c^2}\]in the above expression we have,
\[ = {a^2} - 2ad + {d^2} + 2{b^2} + 2{c^2} - 2{b^2} - 2{c^2}\]
Simplifying we get,
\[ = {a^2} - 2ad + {d^2}\]
So, \[ = {(a - d)^2}\]
Therefore,
\[{(a - d)^2} = {(b - c)^2} + {(c - a)^2} + {(d - b)^2}\] (proved)