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`        Please give answerFind the ap or gp o. ...........    ....  `
6 months ago

Arun
23518 Points
```							a(1/b+1/c) + c(1/a+1/b) =a(c +b)/bc + c(a+b)/ab ={a²(c +b) + c²(a+b)}/abc ={a²c +a²b + c²a+c²b)}/abc ={a²c+ c²a +a²b +c²b)}/abc ={ac(a+c) +a²b +c²b)}/abc  ={ac(2b) +a²b +c²b)}/abc Since a,b,c are in AP hence a+c=2b ={2abc +a²b +c²b)}/abc =b{2ac +a² +c²)}/abc =b{(a+c)²)}/abc ={(a+c)²)}/ac ={(a+c)(a+c))}/ac =(2b)(a+c))}/ac Since a,b,c are in AP hence a+c=2b =(2b)(1/c+1/a) hence a(1/b+1/c), b(1/c+1/a), c(1/a+1/b) are in A.P if a,b,c are in AP
```
6 months ago
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### Course Features

• 101 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions