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Question
For an A.P., show that (m + n)th term + (m – n)th term = 2 × mthterm
Solution
Let a and d be the first term and common difference, respectively.
`\implies` (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 × mth term
Hence proved.
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