Question

John begins to start saving from 1st He saves 1$ on
1st of January, 2$ on 2nd January and 3$ on 3rd of
January. In a similar consecutive fashion, find his
savings by the end of an ordinary year.​

Answer 1

Given :- John begins to start saving from 1st He saves 1$ on 1st of January, 2$ on 2nd January and 3$ on 3rd of January. In a similar consecutive fashion, find his savings by the end of an ordinary year. ?

Solution :-

→ An ordinary year = 365 days.

so,

→ John saves on first day = 1 jan = $ 1 = a1

→ John saves on second day = 2 jan = $ 2 = a2 .

then,

→ d = a2 - a1 = 2 - 1 = $ 1.

then,

→ Sn = (n/2)[2a + (n - 1)d]

→ S(365) = (365/2)[2*1 + (365 - 1)*1]

→ S(365) = (365/2)[2 + 364]

→ S(365) = (365/2) * 366

→ S(365) = $ 66,795 (Ans.)

Hence, John savings by the end of an ordinary year is $ 66,795 .

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