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`        If pth,qth,rth of an AP and GP are both a,b,c respectively show that a^b-c*b^c-a*c^a-b=1`
3 years ago

```							c*b^c-a*c^a-b=(a^b*b^c*c^a)/(a^c*b^a*c^b)=(a/c)^b*(b/a)^c*(c/b)^alet r be the ratio.b=a^r &c=a*r^2.also as in ap 2b=a+c..so if we compute  a^b-=(1/r^2)^b*r^c*r^a=r^(-2b+c+a)=r^0=1
```
3 years ago
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