Simulation · Maths · Class 11

Steady jar against doubling jar

From the lesson Summing n terms of a GP in Sequences and Series. Change the values and watch what happens.

The idea behind it

NCERT §8.4.2

Write Sₙ = a + ar + … + arⁿ⁻¹ and multiply by r: rSₙ = ar + ar² + … + arⁿ. Subtracting, every middle term cancels and (1 − r)Sₙ = a − arⁿ.
So for r ≠ 1, Sₙ = a(1 − rⁿ)/(1 − r) = a(rⁿ − 1)/(r − 1). The first form is handy when |r| < 1, the second when r > 1.
If r = 1 every term equals a and Sₙ = na. The formula must not be used then, because it divides by zero.
The ancestors over 10 generations number 2 + 4 + … + 1024 = 2(2¹⁰ − 1)/(2 − 1) = 2046.
For 3 + 6 + 12 + … to 10 terms, S₁₀ = 3(2¹⁰ − 1) = 3069. For 1 + 1/2 + 1/4 + … to 6 terms, S₆ = (1 − (1/2)⁶)/(1 − 1/2) = 63/32.
Working backwards: 3 + 6 + 12 + … reaches 765 when 3(2ⁿ − 1) = 765, so 2ⁿ = 256 and n = 8.
An A.P. grows by the same amount each time; a G.P. with r > 1 grows by the same factor. Saving ₹100 in week 1 and ₹50 more each week gives a total of ₹4500 after 12 weeks; saving ₹1 in week 1 and doubling each week gives ₹4095 after 12 weeks but ₹8191 after 13, overtaking the ₹5200 of the steady plan.

1 more