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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!
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
- PrakashLv 510 years agoFavorite 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.