if mth term of ap is equal to n times its nth term and m is not equal to n then show m+n th term is 0
Answers
Answer:
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
hence the proof
please mark it as the brainliest
Answer:
We know : an = a +(n-1)d a (m+n) = a + (m+n-1)d (just put m+n in place of n ) --------------(1)
Let the first term and common difference of the A.P. be ‘a’ and ‘d’ respectively. Then,
m th term = a + (m – 1) d and n th term = a + (n-1)d
By the given condition,
↠ m x am = n x an m [a + (m – 1) d] = n [a + (n – 1) d]
↠ ma + m (m – 1) d = na + n (n – 1) d
↠ ma + (m2 -m)d - na - (n2 -n)d = 0 ( taking the Left Hand Side to the other side)
↠ ma -na + (m2 - m)d -( n2-n)d = 0 (re-ordering the terms)
↠ a (m-n) + d (m2-n2-m+n) = 0 (taking 'a ' and 'd ' common)
↠ a (m-n) + d {(m+n)(m-n)-(m-n)} = 0 (a2-b2 identity) Now divide both sides by (m-n)
↠ a (1) + d {(m+n)(1)-(1)} = 0
↠ a + d (m+n-1) = 0 ---------------(2)
• From equation number 1 and 2 :
⇒ a (m+n) = a + (m+n-1)d
And we have shown : a + d (m+n-1) = 0