If m times the mth term of an A.P. is n times its nth term, show that (m n)th term of the A.P. is zero.
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mth term of AP = a + (m-1)d
nth term of AP = a + (n-1)d
Given that
m time mth term = n times nth term
m[a+md-d] = n[a + nd - d]
ma + m²d - md = na + n²d - nd
ma + m²d - md - na - n²d + nd = 0
a(m-n) + d(m²-n²) -d(m-n) = 0
taking (m-n) common and simplifying
a + d(m+n) - d = 0
a + [(m+n) - 1]d = 0
(m+n)th term = 0
nth term of AP = a + (n-1)d
Given that
m time mth term = n times nth term
m[a+md-d] = n[a + nd - d]
ma + m²d - md = na + n²d - nd
ma + m²d - md - na - n²d + nd = 0
a(m-n) + d(m²-n²) -d(m-n) = 0
taking (m-n) common and simplifying
a + d(m+n) - d = 0
a + [(m+n) - 1]d = 0
(m+n)th term = 0
Answered by
1
Answer:
Let the first term of AP = a
common difference = d
We have to show that (m+n)th term is zero or a + (m+n-1)d = 0
mth term = a + (m-1)d
nth term = a + (n-1) d
Given that 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) + (m²-n²)d - (m-n)d = 0
⇒ a(m-n) + {(m-n)(m+n)}d -(m-n)d = 0
⇒ a(m-n) + {(m-n)(m+n) - (m-n)} d = 0
⇒ a(m-n) + (m-n)(m+n -1) d = 0
⇒ (m-n){a + (m+n-1)d} = 0
⇒ a + (m+n -1)d = 0/(m-n)
⇒ a + (m+n -1)d = 0
Proved!
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