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`        Please solve this question as attached in the image. I am not getting the proper solution`
one year ago

1807 Points
```							so we can write fk= Sn – nakor  fk/ak =Sn/ak– n, where Sn = sum of a1, a2, …., anor 1/αk = Sn/ak– nnow, 1/αk – 1/αk-1 = Sn*(1/ak – 1/ak-1)but 1/ak – 1/ak-1= d (a constant coz a1, a2, …., an are in HP)So, 1/αk – 1/αk-1 = Sn*d = constantso α1, α2, α3, …. are in HP tooSo, obviously 2^α1, 2^α2, 2^α3,.... , 2^αn are neither in GP (coz 2^αk /2^ αk-1 is not a constant), nor in AP, and also not in HPso correct option is none of these.
```
one year ago
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• 101 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions