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Question: In an A.P., show that \({a_{m + n}} + {a_{m - n}} = 2{a_m}\)....

In an A.P., show that am+n+am−n=2am{a_{m + n}} + {a_{m - n}} = 2{a_m}.

Explanation

Solution

First find the (m+n)th,(m−n)th{\left( {m + n} \right)^{th}},{\left( {m - n} \right)^{th}} and mth{m^{th}} term of the series and then substitute these values of these terms in the left-hand side of the equation and try to approach the right-hand side of the equation.

Complete answer:
The goal of the problem is to show that am+n+am−n=2am{a_{m + n}} + {a_{m - n}} = 2{a_m} in an A.P. series.
Assume that (a)\left( a \right) denotes any A.P. series, then the nth{n^{th}} term of this series is given as:
an=a+(n−1)d{a_n} = a + \left( {n - 1} \right)d, whereaais the first term of the series and dd be the common difference of the A.P. series.
First, find the (m+n)th,(m−n)th{\left( {m + n} \right)^{th}},{\left( {m - n} \right)^{th}} and mth{m^{th}} term of the series.
So, the (m+n)th{\left( {m + n} \right)^{th}} term of the series is given as:
⇒\Rightarrow am+n=a+(m+n−1)d{a_{m + n}} = a + \left( {m + n - 1} \right)d
⇒\Rightarrow (m−n)th{\left( {m - n} \right)^{th}} term of the series is given as:
⇒\Rightarrow am−n=a+(m−n−1)d{a_{m - n}} = a + \left( {m - n - 1} \right)d
And mth{m^{th}} term is given as:
⇒\Rightarrow am=a+(m−1)d{a_m} = a + \left( {m - 1} \right)d
Now, we have the value of the(m+n)th,(m−n)th{\left( {m + n} \right)^{th}},{\left( {m - n} \right)^{th}}andmth{m^{th}} term of the series.
⇒\Rightarrow am+n=a+(m+n−1)d{a_{m + n}} = a + \left( {m + n - 1} \right)d
⇒\Rightarrow am−n=a+(m−n−1)d{a_{m - n}} = a + \left( {m - n - 1} \right)d
⇒\Rightarrow am=a+(m−1)d{a_m} = a + \left( {m - 1} \right)d
Now, we have to show that:
⇒\Rightarrow am+n+am−n=2am{a_{m + n}} + {a_{m - n}} = 2{a_m}
Then take the left-hand side of the equation and try to approach the right-hand side of the equation:
Left-hand side =am+n+am−n = {a_{m + n}} + {a_{m - n}}
Substitute the values of (m+n)th{\left( {m + n} \right)^{th}} and (m−n)th{\left( {m - n} \right)^{th}} term of the series in the above equation:
Left-hand side =[a+(m+n−1)d]+[a+(m−n−1)d] = \left[ {a + \left( {m + n - 1} \right)d} \right] + \left[ {a + \left( {m - n - 1} \right)d} \right]
Simplify the above equation:
Left-hand side =a+(m+n−1)d+a+(m−n−1)d = a + \left( {m + n - 1} \right)d + a + \left( {m - n - 1} \right)d
Left-hand side =2a+d(m+n−1+m−n−1) = 2a + d\left( {m + n - 1 + m - n - 1} \right)
Left-hand side =2a+d(2m−2) = 2a + d\left( {2m - 2} \right)
Left-hand side =2a+2d(m−1) = 2a + 2d\left( {m - 1} \right)
Left-hand side =2[a+(m−1)d] = 2\left[ {a + \left( {m - 1} \right)d} \right]
We know that the mth{m^{th}} term of the series is given as:
⇒\Rightarrow am=a+(m−1)d{a_m} = a + \left( {m - 1} \right)d
So, the left-hand side becomes:
Left-hand side =2am = 2{a_m}
Left-hand side == Right-hand side

Since the left-hand side of the equation is equal to the right-hand side, therefore the desired result is proved.

Note: The nth{n^{th}} term of the series is given as:
⇒\Rightarrow an=a+(n−1)d{a_n} = a + \left( {n - 1} \right)d
Then for getting the (m+n)th{\left( {m + n} \right)^{th}} term of the series, we have to replace nn by (m+n)\left( {m + n} \right) in the formula:
⇒\Rightarrow am+n=a+(m+n−1)d{a_{m + n}} = a + \left( {m + n - 1} \right)d