Mixed problems

The last part of the chapter mixes the ideas. The first job each time is to spot the progression; the formula comes after.

Sums of squares. A square is never negative, so if several squares add up to 0 (or to something ≤ 0), each square must be 0.

• If (a – 2b)² + (b – 2c)² = 0, then a = 2b and b = 2c.
• So a = 4c, b = 2c, c: each term is half the one before, a G.P. with r = 1/2.

Depreciation is a G.P. with a ratio below 1. A machine worth ₹50000 loses 10% of its value each year, so each year it is multiplied by 0.9. After 3 years: 50000 × 0.9³ = ₹36450.

Interest on what is still owed makes an A.P. A ₹6000 loan is repaid at ₹1000 a month, with 1% interest each month on the amount still owed.

• The interest payments are 60, 50, 40, 30, 20, 10: an A.P. with d = –10.
• Their sum is (6/2)(60 + 10) = 210, so the total paid is 6000 + 210 = ₹6210.

Chain letters. Someone sends a letter to 3 people, and each person who gets one sends it on to 3 more. The rounds have 3, 9, 27, ... letters, a G.P. After 5 rounds: 3(3⁵ – 1)/2 = 363 letters; at ₹2 to post each, that is ₹726.

Sum of a later block of terms. Terms n + 1 to 2n add up to S₂ₙ – Sₙ (all the first 2n terms minus the first n). For a G.P. this works out to rⁿ × Sₙ. Check with 1, 2, 4, 8, 16, 32 and n = 3: the first three add to 7, the next three to 56 = 2³ × 7.

Odd places and all places. If a G.P. has an even number of terms, each term in an even place is r times the term just before it. So the sum of all terms = (1 + r) × (sum of the odd-place terms). If the total is 3 times the odd-place sum, then 1 + r = 3 and r = 2.

Product of terms. The first n terms multiply to aⁿ × r^(0 + 1 + ... + (n – 1)) = aⁿ r^(n(n – 1)/2). For 2, 6, 18, 54: 2⁴ × 3⁶ = 16 × 729 = 11664, which matches 2 × 6 × 18 × 54.

G.P. and A.P. together. 2, 6, 18 are in G.P. Add 4 to the middle number and you get 2, 10, 18, which is an A.P. (difference 8). Questions like this give you one progression and a change that turns it into the other: write both conditions as equations and solve.