Lesson 7 of 10 · 8 min
Problems built on GPs
NCERT §8.4.2
Meera's older cousin reads the notebook and says jar B is just like money earning interest, or a machine losing value. Where else does multiplying by the same number each time turn up?
The lesson in notes
In short
Growth by a fixed factor is a G.P. ₹1000 at 10% interest compounded yearly becomes 1100, 1210, 1331: a G.P. with r = 1.1, so after n years the amount is 1000(1.1)ⁿ.
A culture of 50 bacteria that doubles every hour holds 50 × 2ⁿ after n hours, which is 1600 after 5 hours.
Loss by a fixed fraction is a G.P. too. A machine worth ₹80000 that loses 25% of its value each year is worth 60000, 45000, 33750 after 1, 2, 3 years (r = 3/4).
Three terms in G.P. are best taken as a/r, a, ar, so their product is a³. If the product is 216 then a = 6; if the sum is 19 then 6/r + 6 + 6r = 19, i.e. 6r² − 13r + 6 = 0, so r = 3/2 or 2/3 and the terms are 4, 6, 9 (or 9, 6, 4).
Series such as 5 + 55 + 555 + … are not G.P.s but can be turned into one: 5 + 55 + 555 + … = (5/9)[(10 − 1) + (10² − 1) + … + (10ⁿ − 1)] = (5/81)(10ⁿ⁺¹ − 10) − 5n/9. For n = 3 this gives 615.