if mth term of an A p is eqeal to n times the nth term
find the (m+n)th term of an A.P
Answers
formula
tn=a+(n-1)d
m term of AP
tm=a+(m-1) d...1
ń term of AP
tn=a+(n-1)d...2
from the give condition
m×tm=n×tn
m×[a+(m-1)d]=n×[a+(n-1)d]...[from 1&2]
am+m(m-1)d=an+n(n-1)d
am+m²d-md=an+n²d-nd
am+m²d-n²d-md+nd=0
a(m-n) +d(m²-n²)-d(m-n) =0
(m-n) [a+d(m+n)-d]=0
[a+d(m+n)-d]=0/(m-n)
[a+(m+n-1) d]=0....3
(m+n)th term of the AP
t(m+n)=[a+(m+n-1)d]
t(m+n)=0.....[from 3]
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!