If the m times of mth term is equal to n times of nth term, then show that (m+n)th term is equal to 0.
Answers
Step-by-step explanation:
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
Answer:
0
Step-by-step explanation:
let the first be and common difference is . Given m times the mth term = n times the nth term Recall that the nth term of AP is tm = a + ( n - 1 ) d
mth term of AP =t
m
=a+(m−1)d
⇒mt
m
=nt
n
m[a+(m−1)d]=n[a+(n−1)d]
m[a+(m−1)d]−n[a+(n−1)d]=0
a(m−n)+d[(m+n)(m−n)−(m−n)]=0
(m−n)[a+d((m+n)−1)]=0
a+[(m+n)−1]d=0
But t
m+n
=a+[(m+n)−1]d
∴tm+n = 0