A-Level Sequences and series: worked solution

Question

With the value of r from part (i), find the smallest n for which the sum of all terms beyond the nth term of the geometric progression is under 1% of the sum of its first n terms.

Let the geometric progression have first term a and common ratio r, with r = 3/4 from part (i). Since |r| < 1 the sum to infinity exists, which is what makes 'the sum of all terms beyond the nth term' meaningful. Step 1: write down the two sums being compared. Sum of the first n terms: S_n = a(1 - r^n)/(1 - r) Sum to infinity: S_inf = a/(1 - r) Sum of everything AFTER the nth term = S_inf - S_n = a/(1 - r) - a(1 - r^n)/(1 - r) = a[1 - (1 - r^n)]/(1 - r) = a r^n/(1 - r) Step 2: form the inequality '... is under 1% of the sum of the first n terms'. a r^n/(1 - r) < 0.01 x a(1 - r^n)/(1 - r) Both sides have the positive factor a/(1 - r), so divide it out: r^n < 0.01(1 - r^n) r^n < 0.01 - 0.01 r^n r^n + 0.01 r^n < 0.01 1.01 r^n < 0.01 r^n < 0.01/1.01 = 1/101 Step 3: put r = 3/4 and solve for n. (3/4)^n < 1/101 Take natural logarithms of both sides: n ln(3/4) < ln(1/101) ln(3/4) = -0.28768 is NEGATIVE, so dividing by it reverses the inequality: n > ln(1/101)/ln(3/4) = (-4.6151)/(-0.28768) = 16.04 Step 4: n must be a whole number greater than 16.04, so the least value is n = 17. Check: (3/4)¹⁶ = 0.01004, which is bigger than 1/101 = 0.009901, so n = 16 fails; (3/4)¹⁷ = 0.007528 < 0.009901, so n = 17 works.