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`         if the pth qth and rth terms of an ap be a b and c respectively then show that a(q-r)+b(r-p)+c(p-q)=0`
2 months ago

```							Dear student Let a = first term of the AP.and Let d = common difference of the APNowa = A+(p-1).d.......(i)b = A+(q-1).d.......(ii)c = A+(r-1).d........(iii)Subtracting 2nd from 1st , 3rd from 2nd and 1st from 3rd we geta-b = (p-q).d......(iv)b-c = (q-r).d........(v)c-a = (r-p).d.......(vi)multiply iv,v,vi by c,a,b respectively we havec.(a-b) = c.(p-q).d......(vii)a.(b-c) = a.(q-r).d........(viii)b.(c-a) = b.(r-p).d.......(ix)a(q-r).d+b(r-p).d+c(p-q).d = 0(a(q-r)+b(r-p)+c(p-q)).d = 0Now since d is common difference it should be non zeroHencea(q-r)+b(r-p)+c(p-q)= 0
```
2 months ago
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