If the pth term of an ap is q and qth term is p prove that the nth term is equal to (p+q_n)
Let the AP be
a(n)=a(0)+(n-1)d
then we were given
a(p)=q => a(0)+(p-1)d=q
a(q)=p => a(0)+(q-1)d=p
Subtract one equation from the other
a(0)-a(0)+(p-1-(q-1))d=q-p
simplify
(p-q)d=q-p
solving, we find that
d=-1
substitute d and p in the AP
a(p)=a(0)+(p-1)d
=a(0)+(p-1)(-1)
and also we were given
a(p)=q
therefore
q=a(0)-(p-1)
a(0)=q+p-1
Substitute back in the AP
a(n)=a(0)+(n-1)(-1)
=(q+p-1)+(n-1)(-1)
=q+p-1-n+1
=p+q-n
checking because have concern about form
http://www.mathsisfun.com/algebra/sequences-sums-arithmetic.html
a(n) = a(1) + (n-1)d .......Not a(0)
q = a(1) + (p-1) d
p = a(1) +(q-1) d
p-q = d[ (q-1) - (p-1) ]
p-q = d [q-p]
d = -1 agree
so
a(n) = a(1) +(n-1)(-1)
a(n) = a(1) + 1-n
now need a(1)
a(1) = q -(p-1)(d) = q +(p-1)
a(1) = q +p - 1
so
a(n) = q + p - 1 + 1 - n
a(n) = q + p - n
To prove this statement, we need to use the properties of an arithmetic progression (AP).
Let's start by defining the pth term of the AP as "a" and the common difference as "d". This means that the qth term will be:
q = a + (q-1)d
We are given that the qth term is equal to p. Therefore, we can substitute this into the equation to get:
p = a + (q-1)d
Now, let's find the nth term of the AP, which we'll denote as "Tn". The nth term can be written as:
Tn = a + (n-1)d
We want to prove that Tn = p + qn.
Substituting the values we know, we get:
p + qn = a + (q-1)d + qn
Next, we group the terms with "d" together:
p + qn = a + qd - d + qn
We can rearrange the terms to simplify the equation:
p + qn = a - d + qd + qn
Since a - d represents the first term of the AP (T1), we can rewrite the equation as:
p + qn = T1 + qd + qn
Using the property of an arithmetic progression where the nth term is equal to the first term added to the product of the common difference and (n-1), we can rewrite it as:
p + qn = Tn
Therefore, we have proved that the nth term (Tn) of the AP is equal to p + qn.