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Q.

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

answer is A.

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Detailed Solution

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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