A Level Mathematics: Series, questions with answers

ar² = 50 and ar⁴ = 12.5, so r² = 1/4 and r = 1/2 (positive). Then a = 50/r² = 200.
Sum to infinity = a/(1 − r) = 200/(1/2) = 400.

Divide the two term equations to find r, find a, then use S∞ = a/(1 − r).

The terms are in AP so p² + (p + 6) = 2(4p), giving p² − 7p + 6 = 0, so p = 1 or p = 6.
p = 1: terms 7, 4, 1, d = −3, a = 7 + 27 = 34, u₂₀ = 34 − 57 = −23.
p = 6: terms 12, 24, 36, d = 12, a = 12 − 108 = −96, u₂₀ = −96 + 228 = 132.
Difference = 132 − (−23) = 155.

Use the middle-term property of an AP to find the two values of p, find each progression's 20th term, then subtract.

(a) Sₙ = n/2[4 + (n−1)d]. S₂ = 4 + d, so S₂ − 1 = 3 + d. S₄ = 2(4 + 3d) = 8 + 6d. S₉ = 9/2(4 + 8d) = 18 + 36d.
For an AP: 2S₄ = (S₂ − 1) + S₉ → 2(8 + 6d) = (3 + d) + (18 + 36d) → 16 + 12d = 21 + 37d → −5 = 25d → d = −1/5.
(b) First AP 15th term = 2 + 14d = 2 + 14(−1/5) = −0.8.
Second AP: first term = S₂ − 1 = 3 + d = 2.8, common difference = (8 + 6d) − (3 + d) = 5 + 5d = 4. 15th term = 2.8 + 14(4) = 58.8.
Difference = 58.8 − (−0.8) = 59.6.

Express S₂−1, S₄, S₉ in terms of d, use the arithmetic-sequence condition 2×(middle term)=(sum of outer terms) to find d, then find the 15th term of each progression.

(a) For an AP, common difference d: (4p-1) - 25 = (13-p) - (4p-1)
4p - 26 = 14 - 5p
9p = 40
p = 40/9
d = (4p-1) - 25 = 4(40/9) - 1 - 25 = 160/9 - 26 = (160-234)/9 = -74/9
10th term = 25 + 9d = 25 + 9×(-74/9) = 25 - 74 = -49
(b) For a GP, common ratio r: (4q-1)/25 = (13-q)/(4q-1)
(4q-1)² = 25(13-q)
16q² - 8q + 1 = 325 - 25q
16q² + 17q - 324 = 0
q = [-17 ± √(17² + 4×16×324)]/(2×16) = [-17 ± 145]/32
q = 4 (rejecting the negative root since q is positive)
r = (4(4)-1)/25 = 15/25 = 3/5
Sum to infinity = 25/(1 - 3/5) = 25/(2/5) = 125/2 = 62.5

Use the AP condition (equal common differences) and the GP condition (equal common ratios) to form and solve equations for p and q, then apply the nth term / sum to infinity formulas.

(a) Equal differences: (2p−6) − p²/6 = p − (2p−6).
Multiply by 6: 6(2p−6) − p² = 6p − 6(2p−6) → 12p − 36 − p² = 6p − 12p + 36 → 12p−36−p² = −6p+36.
Rearranging: p² − 18p + 72 = 0 → (p−6)(p−12) = 0 → p = 6 or p = 12.
p = 6 gives common difference 0 (rejected), so p = 12.
(b) With p = 12: first term = 144/6 = 24, second term = 2(12)−6 = 18.
Common ratio r = 18/24 = 3/4.
Sum to infinity = 24/(1 − 3/4) = 24/(1/4) = 96

Set the AP condition to form a quadratic in p, reject the root giving zero common difference, then use S∞ = a/(1−r) for the GP.

Let first term = a, common difference = d.
13th term: a + 12d = 12
Sum of first 30 terms: (30/2)[2a + 29d] = −15 → 2a + 29d = −1
From the first equation: a = 12 − 12d
Substitute: 2(12 − 12d) + 29d = −1
24 − 24d + 29d = −1
24 + 5d = −1
5d = −25
d = −5
a = 12 − 12(−5) = 72
Sum of first 50 terms:
S₅₀ = (50/2)[2a + 49d] = 25[2(72) + 49(−5)] = 25[144 − 245] = 25(−101) = −2525

Turn the two given facts into two simultaneous equations in a and d using the nth-term and sum-of-AP formulae, solve for a and d, then substitute into the sum formula for n = 50.

Coefficient of x² in (1 + (2/p)x)⁵ is ⁵C₂(2/p)² = 10 × 4/p² = 40/p². Coefficient of x² in (1 + px)⁶ is ⁶C₂p² = 15p². So 40/p² + 15p² = 70. Multiplying by p²: 15p⁴ - 70p² + 40 = 0, i.e. 3p⁴ - 14p² + 8 = 0. Let u = p²: 3u² - 14u + 8 = 0 → u = [14 ± √(196 - 96)]/6 = [14 ± 10]/6 → u = 4 or u = 2/3. So p² = 4 → p = ±2, or p² = 2/3 → p = ±√6/3 (AWRT ±0.816).

Find each coefficient using the binomial expansion, form an equation in p, and solve the resulting quartic.

Since consecutive terms have a common ratio r:
(-12)² = (8k)(2k)
144 = 16k²
k² = 9
k = -3 (since k is negative)
r = 2k/(-12) = 2(-3)/(-12) = 0.5
First term a: term 5 = ar⁴ = 8k = -24, so a = -24/(0.5)⁴ = -24/0.0625 = -384
Sum to infinity = a/(1-r) = -384/(1-0.5) = -384/0.5 = -768

Use the fact that (term6)² = term5 × term7 to find k, then find the common ratio and first term, and apply S∞ = a/(1-r).