If the pth term of an AP is q and the qth term is p, prove that its nth term is (p+q-n).
a+(p-1)d = q
a+(q-1)d = p
(p-1)d-q = (q-1)d-p
[(p-1)-(q-1)]d = q-p
(p-q)d = q-p
d = (q-p)/(p-q) = -1
Tn = Tq + (n-q)d
= p + (n-q)(-1)
= p+q-n
thnaks :)
To prove that the nth term of an arithmetic progression (AP) is (p+q-n), where the pth term is q and the qth term is p, we can use the formula for the nth term of an AP.
In an arithmetic progression, the nth term (Tn) can be calculated using the formula:
Tn = a + (n-1)d
where a is the first term of the AP, n is the position of the term we want to find, and d is the common difference.
Let's use this formula to find the expressions for the pth term and the qth term.
For the pth term (Tp):
Tp = a + (p-1)d = q
For the qth term (Tq):
Tq = a + (q-1)d = p
Now, we have two equations:
Tp = q ---(1)
Tq = p ---(2)
To find the values of a and d, we can subtract equation (2) from equation (1):
Tp - Tq = q - p
Substituting the corresponding values:
(a + (p-1)d) - (a + (q-1)d) = q - p
Simplifying the equation:
pd - qd = q - p
Taking d as a common factor:
d(p - q) = q - p
Dividing both sides by (p - q):
d = (q - p)/(p - q)
Note that (p - q) and (q - p) are opposite in sign, so we can rewrite the equation as:
d = -(p - q)/(p - q)
Simplifying further:
d = -1
Since we have found the value of d, we can substitute it into equation (1) to find the value of a:
q = a + (p-1)(-1)
q = a - p + 1
a = q + p - 1
Now, using the formula for the nth term, we can substitute the values of a and d:
Tn = a + (n-1)d
Tn = (q + p - 1) + (n-1)(-1)
Tn = q + p - 1 - n + 1
Tn = q + p - n
Therefore, we have proved that the nth term of the AP is (p+q-n).