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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

```
9 years ago

```							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
```
9 years ago
```							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

```
9 years ago 605 Points
```							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
```
4 months ago
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