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# 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.  Verified
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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 ‘a’ be the first term and ‘d’ be the common difference of the given A.P.

From general formula of A.P. we have ‘n’ 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)

Let’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 ‘2’ 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) ${T_n} = a + (n - 1)d$