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Geometrical Progression(GP) Math!!?

The sum of the first n terms of a progression is 3 power n-1 .Prove that the progression is a Geometrical Progression(GP)

Do you agree that;

Maths is sometimes just awesome and sometimes........

-If you like math mark the question interesting or write a note along with the answer you provide..:)

Thanks!

Update:

Also help with this!

In how many years would one triple one's investment if sh.250,000 is invested at 8% compounded annually?

Thanks once again!!

1 Answer

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  • 10 years ago
    Favorite Answer

    1. Sum of 1st n terms of a geometric progression = a*(r^n - 1) / (r - 1)

    and you are saying, it is equal to 3^n - 1

    Comparing, we see that if a = 2, and r = 3, we get

    a*(r^n - 1) / (r - 1) = 2*(3^n - 1) / (3 - 1) = 3^n - 1, and that is what you wanted.

    Hope it helps.

    2. In how many years would one triple one's investment if sh.250,000 is invested at 8% compounded annually?

    Amount A = P*(1+r in decimal)^n, number of years

    Therefore 3*250,000 = 250,000*(1+0.08)^n

    or 3 = 1.08^n

    Taking logs,

    log 3 = n*log 1.08

    0.477121254719662 = n*0.0334237554869497

    which gives n = 14.27491459 = approx 14 years 4 months

    Hope it helps too.

    Yes, young lady, the question is interesting.

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