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)

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.

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