Question

# Show that the sum of ${(m + n)^{th}}$ and ${(m - n)^{th}}$ term of an A.P. is equal to twice the ${m^{th}}$ term.

Hint: Any arithmetic progression follows the sequence of a, a + d, a + 2d, .... + a +(n-1)d . Sum of these all terms will be the sum of that arithmetic progression.

Let â€˜aâ€™ be the first term and â€˜dâ€™ be the common difference of the given A.P.

From general formula of A.P. we have â€˜nâ€™ th term A.P. is ${T_n} = a + (n - 1)d$
The m th term is ${T_m} = a + (m - 1)d$ ... (1)

Then ${(m + n)^{th}}$ term $\Rightarrow {T_{m + n}} = a + (m + n - 1)d$ ... (2)

${(m - n)^{th}}$ term $\Rightarrow {T_{m - n}} = a + (m - n - 1)d$ .... (3)

Letâ€™s find the sum of ${(m + n)^{th}}$ and ${(m - n)^{th}}$ terms, adding equation (2) and (3)

$\Rightarrow {T_{m + n}} + {T_{m - n}} = \left( {a + (m + n - 1)d} \right) + \left( {a + (m - n - 1)d} \right)$

$\Rightarrow {T_{m + n}} + {T_{m - n}} = \left( {a + (m + n - 1)d} \right) + \left( {a + (m - n - 1)d} \right)$

$\Rightarrow {T_{m + n}} + {T_{m - n}} = \left[ {2a + (m + n - 1 + m - n - 1)d} \right]$

$\Rightarrow {T_{m + n}} + {T_{m - n}} = \left[ {2a + (2m - 2)d} \right]$

Taking out â€˜2â€™ from RHS
$\Rightarrow {T_{m + n}} + {T_{m - n}} = 2\left[ {a + (m - 1)d} \right]$

From equation (1), we can substitute
$\Rightarrow {T_{m + n}} + {T_{m - n}} = 2{T_m}$

Hence proved.

$\therefore$The sum of ${(m + n)^{th}}$ and ${(m - n)^{th}}$ term of an A.P. is equal to twice the ${m^{th}}$ term.

Note: Arithmetic progression (A.P.) is a sequence of numbers, in which the difference between consequent numbers is a fixed number (common difference) throughout the sequence. ${n^{th}}$ term of A.P. depends on first term(a) and the common difference(d) ${T_n} = a + (n - 1)d$