# 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.

• 36

a + (p - 1)d = q

a + (q - 1)d = p

solving, a =  q(q - 1) - p(p - 1)/(q - p) = q2 - p2 + p - q/ (q - p) = q +p - 1

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

nth term = a + (n - 1)d

= q + p - 1 - n + 1

= q + p - n

thumbs up plzzzzz

• 66

thanks

• -23

thanks dear thanks a lot

• -23

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,

tn = a + (n-1)d

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

tn =(p+q-1) + (n-1)-1

=p+q-1-n+1

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

good luck... :)

• 353

Good

• -31
In an ap proove that d=Sn-2Sn-1+Sn-2
• -20
i cant understand
• -23
Its really interesting.
• -25
Let first term be a and common difference be d.

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
• -4
Toes up!! • -6
thanks vartika.
• -18
Rahe de
• -18
How r u?
• -24
thank u vartika dear
• -17
how do u ask question in meritnation. i know how to answer but idk how to ask pls help. i will be following dis question.
• -10
Hope it will help!!!👍👍👍 • 35
• -12
Thamk you varti ka😘🙏🙏🙏
• -8
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, tn = a + (n-1)d by putting the value of a and d,we get, tn =(p+q-1) + (n-1)-1 =p+q-1-n+1 I hope this is helpful for you =(p+q-n).....proved!
• 4
Here it is⬇️⬇️ • 13
Let first term be a and common difference be d.
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
• 4
Divide the polynomial 2x^2-6x^3+7x^2-4x-2 by the polynomial 2x^2-2x+1 and verify (by) division algorithm. • 6 • 4 • 0 • 1
If u find it useful then press the smiley.. 😊 • 0
i hope its helped u • 0
its help u • 0 • 0
it is clearly given in the r d sharma guyss
• 0
Ang Daan ka mahatva batate Hue Apne Mata Pita ko Patra likhe in Hindi mai
• 0 • 0 • 0
Solution • 0 • 0
Sorry i dont know
• 1 • 0 • 0 • 0
Consider the sum of three no.s 14,27and13we may do the sum in 2 ways
• 0
ufuc8wg8wg8ggigd8hxa9gxq8
• 0 • 0 • 0
The 7 th of an AP is 4 times its 2 nd term and 12 th is 2 more than 3 times of its 4 th term find AP
• 0
Hope this will be helpful for you • 0
Given,
• 0  