Question

Let the sequence $ {a_1},{a_2},{a_3},...{a_{2n}} $ form an A.P, then $ {a_1}^2 - {a_2}^2 + a_3^2 - ... + a_{2n - 1}^2 - a_{2n}^2 = $
A. $ \dfrac{{n(a_1^2 - a_{2n}^2)}}{{2n - 1}} $
B. $ \dfrac{{2n(a_{2n}^2 - a_1^2)}}{{n - 1}} $
C. $ \dfrac{{n(a_1^2 + a_{2n}^2)}}{{n + 1}} $
D.None of these

Answer 1

Hint: In this question we have been given
  $ {a_1},{a_2},{a_3},...{a_{2n}} $ form an A.P . So we will first calculate the common difference i.e.
  $ d $ and then we will break the term
 $ {a_1}^2 - {a_2}^2 + a_3^2 - ... + a_{2n - 1}^2 - a_{2n}^2 $ in the product of two terms and then we will simplify it.

Complete step-by-step answer:
Let us first understand what an arithmetic progression is.
We know that an arithmetic progression is a sequence where each term is found by adding a fixed amount on the previous term. The fixed amount is called common difference.
The general form of an arithmetic progression is
 $ {a_1},{a_2},{a_3},...{a_n} $ .
The formula of an arithmetic progression is
 $ {a_n} = a + (n - 1)d $
From the question, we have been given
 $ {a_1},{a_2},{a_3},...{a_{2n}} $ .
We can calculate the common difference i.e.
 $ d = {a_2} - {a_1} $
We can also write this by interchanging the terms without changing its true form as
  $ {a_1} - {a_2} = - d $
Similarly we can write
 $ d = {a_4} - {a_3} $ , Or it can be written as
 $ {a_3} - {a_4} = - d $
We can write them all together as
 $ d = {a_2} - {a_1} = {a_4} - {a_3} = {a_{2n}} - {a_{2n - 1}} $ .
Now we have another term in the question $ {a_1}^2 - {a_2}^2 + a_3^2 - ... + a_{2n - 1}^2 - a_{2n}^2 $ .
We can apply the formula here:
 $ {a^2} - {b^2} = (a + b)(a - b) $ .
So we can write the above expression as:
 $ ({a_1} - {a_2})({a_1} + {a_2})({a_3} - {a_4})({a_3} + {a_4}) + ... + ({a_{2n - 1}} - {a_{2n}})({a_{2n - 1}} + {a_{2n}}) $
From the above we can put the value of $ - d $ , it gives:
 $ ( - d)({a_1} + {a_2})( - d)({a_3} + {a_4}) + ... + ( - d)({a_{2n - 1}} + {a_{2n}}) $
By taking the common factor out and adding the terms we have:
 $ - d({a_1} + {a_2} + {a_3} + {a_4} + ... + {a_{2n}}) $
We can see that again we have an A.P, where the first term is
 $ a = {a_1} $ , number of terms i.e.
 $ n = 2n $ and the last term is $ {a_{2n}} $ .
So by applying the formula of sum of an A.P, we have :
 $ - d\left( {\dfrac{{2n}}{2}({a_1} + {a_{2n}})} \right) $
On simplifying the value, we have;
 $ - nd({a_1} + {a_{2n}}) $
We can apply this formula i.e.
 $ {a_n} = a + (n - 1)d $ , we can calculate the value of $ {a_{2n}} $ .
So we have $ 2n $ in place of $ n $ terms, it gives
 $ {a_{2n}} = {a_1} + (2n - 1)d $
We can simplify this expression:
 $ {a_{2n}} - {a_1} = (2n - 1)d $
We can calculate the common difference i.e.
  $ d = \dfrac{{{a_{2n}} - {a_1}}}{{2n - 1}} $
Or, it can be written as
 $ - d = \dfrac{{{a_1} - {a_{2n}}_{}}}{{2n - 1}} $
We will now put this value in the expression and it gives us :
 $ - n \times \left( { - \dfrac{{({a_1} - {a_{2n}})}}{{2n - 1}}} \right)({a_1} + {a_{2n}}) $
On simplifying it gives us value:
 $ \dfrac{{n({a_1} - {a_{2n}})}}{{2n - 1}}({a_1} + {a_{2n}}) $
Again by applying the formula
  $ {a^2} - {b^2} = (a + b)(a - b) $ , we can write the above expression as :
 $ \dfrac{n}{{2n - 1}}(a_1^2 - a_{2n}^2) $
Hence the correct option is (a) $ \dfrac{n}{{2n - 1}}(a_1^2 - a_{2n}^2) $
So, the correct answer is “Option a”.

Note: We should note that in arithmetic progression, we have been given the first term and the last term, then the formula of sum of the terms is
  $ {S_n} = \dfrac{n}{2}\left( {a + l} \right) $ , where $ a $ is the first term and
 $ l $ is the last term. We have applied this formula in the above solution