Question

Let \[{a_n}\]be the \[n^{th}\] term of an AP. If \[\sum\limits_{r = 1}^{100} {{a_{2r}} = \alpha } \] and If \[\sum\limits_{r = 1}^{100} {{a_{2r - 1}}} = \beta \] , then the common difference of an AP is
(a) \[\left( {\alpha - \beta } \right)/200\]
(b) \[\left( {\alpha - \beta } \right)/100\]
(c) \[(\alpha - \beta )\]
(d) \[(\beta - \alpha )\]

Answer 1

Hint:In this question, \[{a_n}\]is the \[n^{th}\] term of the arithmetic progression.we will find the of ‘d’, expand the given summations by substituting the values of r as \[r = 1,\,r = 2,r = 3...\] up to 100 terms. Now, Subtract the expanded equations and use the formula to find the common difference ‘d’ which is given by \[d = {a_n} - {a_{n - 1}}\]

Formula Used:
 \[d = {a_n} - {a_{n - 1}}\]

Complete step by step Solution:
Given:
\[\displaystyle\sum\limits_{r=1}^{100} a_{2r}=\alpha\]
Expanding the above summation by substituting \[r = 1,\,r = 2,r = 3...\] , we get
\[{a_2} + {a_4} + {a_6} + \ldots + {a_{200}} = \alpha \]--- equation (1)
\[\sum\limits_{r = 1}^{100} {{a_{2r - 1}}} = \beta \]
Expanding the above summation by substituting\[r = 1,\,r = 2,r = 3...\], we get
\[{a_1} + {a_3} + {a_5} + \ldots + {a_{199}} = \beta \] ---- equation (2)
Now, Subtracting equation (1) from equation (2)
\[({a_2} + {a_4} + {a_6} + \ldots + {a_{200}}) - ({a_1} + {a_3} + {a_5} + \ldots + {a_{199}}) = \alpha - \beta \]
Simplify the above equation and solve for common difference ‘d’
\[\left( {{a_2} - {a_1}} \right) + \left( {{a_4} - {a_3}} \right) + \left( {{a_6} - {a_5}} \right) + \ldots .\left( {{a_{200}} - {a_{199}}} \right) = \alpha - \beta \] -- equation (3)
The common difference d in the arithmetic progression is given by \[d = {a_n} - {a_{n - 1}}\]
Therefore, equation (3) becomes
\[d + d + d + ......100 times = (\alpha - \beta )\]
Since r varies from 1 to 100 in the given question, the common difference ‘d’ adds up to 100 terms. Thus the equation obtained is
\[100d = \alpha - \beta \]
Solving for common difference ‘d’
Therefore, \[d = \dfrac{{(\alpha - \beta )}}{{100}}\]
Hence, the correct option is b.

Note: The value of each next number in an arithmetic progression or sequence differs from the previous one. \[{a_n}\] is the nth term in the sequence, and \[{a_{\left( {n - 1} \right)}}\] is the preceding term ..An arithmetic progression is a sequence of numbers such that the difference between consecutive numbers or terms is constant. This constant number is defined as a common difference and is denoted by d.