If m times the mth term of an A.P. is equal to n times the nth term, find its (m+n)th term.
Answers
Step-by-step explanation:
mth term = a+(m-1)d
m times of mth term = [a+(m-1)d]m
nth term = a+(n-1)d
n times of nth term = [a+(n-1)d]n
Given, [a+(m-1)d]m = [a+(n-1)d]n
= (a+md-d)m = (a+nd-d)n
= am+m^2d-dm = an+n^2d-dn
= am+m^2d-dm -an-n^2d+dn
= m(a+md-d) +n(-a-n+d)
(m+n)th term = a+(m-1)d + a+(n-1)d
= 2a+md-d+nd-d
= 2a+md+nd-2d
= 2(a-d) + d(m+n)
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!