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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 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

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thanks

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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... :)

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Good

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In an ap proove that d=Sn-2Sn-1+Sn-2
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i cant understand
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Its really interesting.
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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
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Toes up!!

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thanks vartika. 
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Rahe de
  • -18
How r u?
  • -24
thank u vartika dear
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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.
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Hope it will help!!!👍👍👍

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Funny answer
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Thamk you varti ka😘🙏🙏🙏
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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, 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!
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Here it is⬇️⬇️

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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
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