If the pth term of an A. P. is q and the qth is p, then prove that it's nth term is( p+q-n)
Answers
Step-by-step explanation:
Given :-
The pth term of an A. P. is q and the qth is p.
To find :-
Prove that it's nth term is( p+q-n) .
Solution :-
Let the first term of an AP = a
Let the common difference of the AP = d
We know that
the nth term of an AP = a+(n-1)d ---------(1)
Given that
pth term of the AP = ap = q
=>q = a+(p-1)d -------------(2)
Given that
qth term of the AP = aq = p
=>p= a+(q-1)d --------------(3)
On subtracting (2) from (3)
p = a+(q-1)d
q = a+(p-1)d
(-)
_________________
p-q = 0 + (q-1)d - (p-1)d
__________________
=> p-q = (q-1)d - (p-1)d
=> p-q = qd - d - pd + d
=> p-q = qd-pd+(d-d)
=> p-q = qd-pd
=> p-q = d(q-p)
=> p-q = -(p-q) d
=> d = (p-q)/-(p-q)
=> d = -1
Common difference = -1
On Substituting the value of d in (2) then
q = a+(p-1)(-1)
=> q = a -p+1
=> a = q+p-1
Now
nth term = an
an = a+(n-1)d
=> an = (q+p-1)+(n-1)(-1)
=>an = q+p-1-n+1
=>an = p+q-n+1-1
=> an = p+q-n
Hence, Proved.
Answer:-
nth term of the given AP for the given problem is
(p+q-n)
Used formulae:-
- the nth term of an AP = a+(n-1)d
Where,
- a= First term
- d = Common difference
- n= number of terms