If the mth term of an A. P. is
and nth term is
then show that its (mn)th term is 1.
Given: ![]()
Now, am = a + (m – 1)d
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⇒ an + n(m – 1)d = 1
⇒ an + mnd – nd = 1 …(i)
![]()
![]()
⇒ am + mnd – md = 1 …(ii)
From eq. (i) and (ii), we get
an + mnd – nd = am + mnd – md
⇒ an – am – nd + md = 0
⇒ a(n – m) – d (n – m) = 0
⇒ a = d
Now, putting the value of a in eq. (i), we get
dn + mnd – nd = 1
⇒ mnd = 1
![]()
Hence, ![]()
Now, (mn)th terms of AP is
amn = a + (mn – 1)d
![]()
![]()
⇒ amn = 1
Hence Proved
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