if the pth term of an A.P is q and the qth term is p, prove that its nth term is (p+q-n)

Let the A.P. be given by

a, a+d, a+2d, a+3d,.....

q = a + (p -1)d ... ( 1 )

p = a + (q -1)d ... ( 2 )

Subtracting ( 2 ) from ( 1 ),

q - p = (p - q)d => d = -1

Putting d = -1 in ( 1 ), a = p + q - 1

nte term = a + (n - 1)d

Substituting the values of a and d,

nth term = p + q - 1 + (n -1)(-1) = p + q - n.

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according to question,

a + (p-1)d = q ....1

and a+(q-1)d = p ....2

by subtracting 2 from 1 we get,

(p-1)d - (q-1)d = q-p

d((p-1)-(q-1)) = (q-p)

d(p-1-q+1)= (q-p)

d(p-q) = (q-p)

so,d = -1 ......3

putting value of d in 1,we get,

a+(p-1) -1=q

a-p+1 = q

so,a= (q+p-1) ......4

now,we know that,

t_{n} = a + (n-1)d

by putting the value of a and d,we get,

t_{n} =(p+q-1) + (n-1)-1

=p+q-1-n+1

=(p+q-n).....proved!

good luck... :)

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pth term = a + (p-1)d = q -----(i)

qth term = a?+ (q-1)d = p -----(ii)

(ii) - (i) gives

(q - p)d = p - q or d = -1

(i) gives? a = q - (p-1)d = q - (p-1)(-1) = q + p - 1

Hence,

nth term = a + (n-1)d = q + p -1 + (n-1)(-1) = q + p -1 + n +1 = p + q - n

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pth term = a + (p-1)d = q -----(i)

qth term = a + (q-1)d = p -----(ii)

(ii) - (i) gives

(q - p)d = p - q or d = -1

(i) gives a = q - (p-1)d = q - (p-1)(-1) = q + p - 1

Hence,

nth term = a + (n-1)d = q + p -1 + (n-1)(-1) = q + p -1 + n +1 = p + q - n

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