An algebra problem by Jimmy PrevailLone
Given arithmetic sequence a m and a n If a m = n and a n = m Find a m + n
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2 solutions
This is not correct enough. Your answer assumes that m = n yet there is no statement saying that. I think the answer should be:
a m + n = a m + n d = a n + m d ⇒ n + n d = m + m d ⇒ d = − 1 ( m = n )
Or m = n ( d ∈ R )
If d = − 1 , then a m + n = 0 ;
If m = n , then a m + n = m + n .
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In fact, when m = n , you cannot tell what a m + n is.
dat's so simple !! The ans is 0
Since a(n)=m and a(m)=n, the common difference must be -1. Then, a(n+m)= a(n)-(m)(-1)=0
this question simply implies that this arithmatic series progresses with a negative common difference such that the term m+n = 0 and all terms > m+n ( starting with m+n+1) are negative
It is an arithmetic progression. So it will have initial value 'a' & common difference 'd'. Our first step is to find these.
a(x) = a + (x-1)d
From the given values,
n = a + (m-1)d &
m = a + (n-1)d
Solving, we get
a = n+m-1 d = -1
So, a(m+n) = a + (m+n-1)d
Substituting 'a' & 'd' we get