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If m times the mth term of an A.P. is equal to n times nth term then show that the (m + n)th term of the A.P. is zero.

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Sol.      Let a be the first term and be the common difference.
∴                                                                mtm  =  ntn                      [Given]
∴                                               m[a + (m – 1)d]  = n[a + (n – 1)d]
∴                                               am + m2d – md = an + n2d – nd
∴                     am – an + m2d – n2d – md + nd = 0
∴                   a(m – n) + d(m2 – n2) – d(m – n)  = 0
∴          a(m – n) + d(m + n) (m – n) – d (m – n) = 0
∴                                (m – n) [a + d(m + n) – d] = 0                         [Taking (m – n) common out of all terms]
∴                                (m – n) [a + (m + n – 1)d] = 0
∴                m – n = 0    or        a + (m + n – 1)d = 0
tm and tn are two different terms
∴         m ≠ n ∴ m – n ≠ 0
∴         a + (m + n – 1)d = 0                            ...(I)
Now, t(m+n) = a + (m + n – 1)d         ...(II)
∴                  t(m+n) = 0                                    [From (I) and (II)]

Arithmetic Progression September 28 , 2018 0 Comments 379 views