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1+4/5+7/5^2+10/5^3+......anant pado ka yog hai Sin^2(π/8+A/2) -sin^2(π/8-A/2)

1+4/5+7/5^2+10/5^3+......anant pado ka yog hai
Sin^2(π/8+A/2) -sin^2(π/8-A/2) 

Grade:12

2 Answers

Aditya Gupta
2081 Points
4 years ago
hello ravi, note that the given series is an infinite AGP (arithmetico geometric progression) with common diff d= 3, common ratio r= 1/5, a= 1. we can directly use the formula for the sum of an infinite AGP:
S= a/(1 – r) + dr/(1 – r)^2
putting the values, we have
S= 35/16
now i dont understand what you mean by sin^2(π/8+A/2) -sin^2(π/8-A/2) ?? what is A here. kindly clarify.
kindly approve :)
Arun
25750 Points
4 years ago
Dear student
 
In 2nd question
 
it is formula of sin²x - sin²y = sin (x+y) sin (x-y)
Hence
 = Sin (π/8 +A/2 + π/8 -A/2) sin (π/8 + A/2 - (π/8 -A/2))
 = Sin (π/4) sin (A)
 = (1/√2) sinA

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