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1.IF Pth term of an A.P is q and qth tem of an AP is p then prove that (p+q) term is 0
let a & d are the first term & common differnece respectively then pth term = a+(p-1)d = q ...............1 qth term = a+(q-1)d = p ...............2 1-2 q-p = (p-q)d d=-1 now (p+q)th term = a + (p+q-1)d putting value of a from eq 1 (p+q)th term = q - (p-1)d + (p+q-1)d (d =-1) =0
let a & d are the first term & common differnece respectively then
pth term = a+(p-1)d = q ...............1
qth term = a+(q-1)d = p ...............2
1-2
q-p = (p-q)d
d=-1
now
(p+q)th term = a + (p+q-1)d
putting value of a from eq 1
(p+q)th term = q - (p-1)d + (p+q-1)d (d =-1)
=0
The pth term of an AP is = a+(p-1)d where a is the first term of the AP and d is the common difference as pth term is equal to q therefore q=a+(p-1)d .....1 similarly qth term is equal to p p=a+(q-1)d .......2 solve the two equations to get the value of a and d. It will come a=p+q-1 d=-1 now (p+q)th term is =a+(p+q-1)d put the values of a and d =(p+q-1)+(p+q-1)(-1) =(p+q-1)-(p+q-1) =0
The pth term of an AP is = a+(p-1)d
where a is the first term of the AP and d is the common difference
as pth term is equal to q therefore q=a+(p-1)d .....1
similarly qth term is equal to p
p=a+(q-1)d .......2
solve the two equations to get the value of a and d. It will come
a=p+q-1
now (p+q)th term is =a+(p+q-1)d
put the values of a and d
=(p+q-1)+(p+q-1)(-1)
=(p+q-1)-(p+q-1)
Dear student,Please find the attached solution to your problem. Let the first term of A.P. = a and common difference = dpth term → a + (p-1)d = q – (1)qth term → a + (q-1)d = p – (2)eq(1) – eq(2)(p-q)d = q-por, d = -1putting d = -1 in eq(1) we get, a = p+q-1Now, (p+q)th term → a + (p+q-1)d = p+q-1 - (p+q-1) = 0 Hope it helps.Thanks and regards,Kushagra
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