Question
For an A.P., show that:
$(m + n)^{th} ~term + (m - n)^{th} ~term = 2 \times m^{th} ~term$
$(m + n)^{th} ~term + (m - n)^{th} ~term = 2 \times m^{th} ~term$
✓
Answer
Let a and d be the first term and common difference respectively.
$\Rightarrow (m + n)^{th} term = a + (m + n - 1)d$ …. (i) and
$(m - n)^{th} term = a + (m - n - 1)d$ …. (ii)
From (i) + (ii), we get
$(m + n)^{th} term + (m - n)^{th} term$
$= a + (m + n - 1)d + a + (m - n - 1)d$
$= a + md + nd - d + a + md - nd - d$
$= 2a + 2md - 2d$
$= 2a + (m - 1)2d$
$= 2[ a + (m - 1)d]$
$= 2 \times m^{th}term$
Hence proved.
$\Rightarrow (m + n)^{th} term = a + (m + n - 1)d$ …. (i) and
$(m - n)^{th} term = a + (m - n - 1)d$ …. (ii)
From (i) + (ii), we get
$(m + n)^{th} term + (m - n)^{th} term$
$= a + (m + n - 1)d + a + (m - n - 1)d$
$= a + md + nd - d + a + md - nd - d$
$= 2a + 2md - 2d$
$= 2a + (m - 1)2d$
$= 2[ a + (m - 1)d]$
$= 2 \times m^{th}term$
Hence proved.
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