Question

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

Answer 1

Hint: First find the ${\left( {m + n} \right)^{th}},{\left( {m - n} \right)^{th}}$ and ${m^{th}}$ 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 ${a_{m + n}} + {a_{m - n}} = 2{a_m}$ in an A.P. series.
Assume that $\left( a \right)$ denotes any A.P. series, then the ${n^{th}}$ term of this series is given as:
${a_n} = a + \left( {n - 1} \right)d$, where$a$is the first term of the series and $d$ be the common difference of the A.P. series.
First, find the ${\left( {m + n} \right)^{th}},{\left( {m - n} \right)^{th}}$ and ${m^{th}}$ term of the series.
So, the ${\left( {m + n} \right)^{th}}$ term of the series is given as:
$ \Rightarrow $ ${a_{m + n}} = a + \left( {m + n - 1} \right)d$
$ \Rightarrow $ ${\left( {m - n} \right)^{th}}$ term of the series is given as:
$ \Rightarrow $ ${a_{m - n}} = a + \left( {m - n - 1} \right)d$
And ${m^{th}}$ term is given as:
$ \Rightarrow $ ${a_m} = a + \left( {m - 1} \right)d$
Now, we have the value of the${\left( {m + n} \right)^{th}},{\left( {m - n} \right)^{th}}$and${m^{th}}$ term of the series.
$ \Rightarrow $ ${a_{m + n}} = a + \left( {m + n - 1} \right)d$
$ \Rightarrow $ ${a_{m - n}} = a + \left( {m - n - 1} \right)d$
$ \Rightarrow $ ${a_m} = a + \left( {m - 1} \right)d$
Now, we have to show that:
$ \Rightarrow $ ${a_{m + n}} + {a_{m - n}} = 2{a_m}$
Then take the left-hand side of the equation and try to approach the right-hand side of the equation:
Left-hand side $ = {a_{m + n}} + {a_{m - n}}$
Substitute the values of ${\left( {m + n} \right)^{th}}$ and ${\left( {m - n} \right)^{th}}$ term of the series in the above equation:
Left-hand side $ = \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 + \left( {m + n - 1} \right)d + a + \left( {m - n - 1} \right)d$
Left-hand side $ = 2a + d\left( {m + n - 1 + m - n - 1} \right)$
Left-hand side $ = 2a + d\left( {2m - 2} \right)$
Left-hand side $ = 2a + 2d\left( {m - 1} \right)$
Left-hand side $ = 2\left[ {a + \left( {m - 1} \right)d} \right]$
We know that the ${m^{th}}$ term of the series is given as:
$ \Rightarrow $ ${a_m} = a + \left( {m - 1} \right)d$
So, the left-hand side becomes:
Left-hand side $ = 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 ${n^{th}}$ term of the series is given as:
$ \Rightarrow $ ${a_n} = a + \left( {n - 1} \right)d$
Then for getting the ${\left( {m + n} \right)^{th}}$ term of the series, we have to replace $n$ by $\left( {m + n} \right)$ in the formula:
$ \Rightarrow $ ${a_{m + n}} = a + \left( {m + n - 1} \right)d$