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 If it is given that 114+124+134+ is π490 then the value of 114+134+154+

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a
π496
b
π445
c
89π490
d
π696

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detailed solution

Correct option is A

let x=114+134+154+...∞given equation is =π490=114+134+154+…+124114+124+134+…π490=x+116⋅π490x=π4901−116=π490×1516=π496


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