Home / CIE A level -Pure Mathematics 1 : Topic : 1.6 Series: arithmetic and geometric progressions : Exam Questions Paper 1

Question

The first term of a convergent geometric progression is 10. The sum of the first 4 terms of the progression is \( p \) and the sum of the first 8 terms of the progression is \( q \). It is given that \( \frac{q}{p} = \frac{17}{16} \). Find the two possible values of the sum to infinity.

▶️ Answer/Explanation

Solution

For a geometric series with first term \( a = 10 \) and common ratio \( r \), the sum of the first \( n \) terms is:
\( S_n = a \frac{1 – r^n}{1 – r} \)
Thus:
– Sum of first 4 terms: \( p = S_4 = 10 \frac{1 – r^4}{1 – r} \)
– Sum of first 8 terms: \( q = S_8 = 10 \frac{1 – r^8}{1 – r} \)
Given: \( \frac{q}{p} = \frac{17}{16} \)
\( \frac{q}{p} = \frac{10 \frac{1 – r^8}{1 – r}}{10 \frac{1 – r^4}{1 – r}} = \frac{1 – r^8}{1 – r^4} \)
Since \( 1 – r^8 = (1 – r^4)(1 + r^4) \):
\( \frac{1 – r^8}{1 – r^4} = \frac{(1 – r^4)(1 + r^4)}{1 – r^4} = 1 + r^4 \)
Thus: \( 1 + r^4 = \frac{17}{16} \)
\( r^4 = \frac{17}{16} – 1 = \frac{1}{16} \)
\( r = \pm \sqrt[4]{\frac{1}{16}} = \pm \frac{1}{2} \)
Since the series converges, \( |r| < 1 \), so \( r = \frac{1}{2} \) and \( r = -\frac{1}{2} \) are valid.
Sum to infinity: \( S_\infty = \frac{a}{1 – r} \)
– For \( r = \frac{1}{2} \): \( S_\infty = \frac{10}{1 – \frac{1}{2}} = \frac{10}{\frac{1}{2}} = 20 \)
– For \( r = -\frac{1}{2} \): \( S_\infty = \frac{10}{1 – (-\frac{1}{2})} = \frac{10}{\frac{3}{2}} = \frac{20}{3} \)
Answer: \( 20, \frac{20}{3} \)

Question

(a) Find the coefficients of \(x^3\) and \(x^4\) in the expansion of \((3 – ax)^5\) in terms of \(a\).
(b) Given the coefficient of \(x^4\) in \((ax + 7)(3 – ax)^5\) is 240, find the positive value of \(a\).

▶️ Answer/Explanation

Solution

(a) For \((3 – ax)^5 = \sum_{r=0}^5 \binom{5}{r} 3^{5-r} (-a)^r x^r\):
Coefficient of \(x^3\) (r=3): \(\binom{5}{3} \cdot 3^2 \cdot (-a)^3 = 10 \cdot 9 \cdot (-a^3) = -90a^3\)
Coefficient of \(x^4\) (r=4): \(\binom{5}{4} \cdot 3^1 \cdot a^4 = 5 \cdot 3 \cdot a^4 = 15a^4\)
Answer (a): \(-90a^3, 15a^4\)

(b) Find the coefficient of \(x^4\) in \((ax + 7)(3 – ax)^5\), given it equals 240.
Steps:
1. Identify terms in \((ax + 7)(3 – ax)^5\) that yield \(x^4\):
   * \(ax \cdot (\text{coefficient of } x^3 \text{ in } (3 – ax)^5)\): gives \(x^{1+3} = x^4\).
   * \(7 \cdot (\text{coefficient of } x^4 \text{ in } (3 – ax)^5)\): gives \(x^4\).
2. From (a), coefficient of \(x^3\) is \(-90a^3\), and \(x^4\) is \(15a^4\).
3. Calculate contributions:
   * For \(ax \cdot (-90a^3 x^3)\): coefficient is \(a \cdot (-90a^3) = -90a^4\).
   * For \(7 \cdot (15a^4 x^4)\): coefficient is \(7 \cdot 15a^4 = 105a^4\).
4. Sum the coefficients: \(-90a^4 + 105a^4 = 15a^4\).
5. Set equal to 240: \(15a^4 = 240\).
6. Solve: \(a^4 = \frac{240}{15} = 16 \implies a = \sqrt[4]{16} = 2\) (positive).
Answer (b): \(2\)
Final Answer: \((a) -90a^3, 15a^4; (b) 2\)

Question

An arithmetic progression has fourth term 15 and eighth term 25. Find the 30th term of the progression.

▶️ Answer/Explanation

Solution

Let the first term be \(a\) and common difference be \(d\).
Given:
Fourth term: \(a + 3d = 15\) (1)
Eighth term: \(a + 7d = 25\) (2)
Subtract (1) from (2):
\((a + 7d) – (a + 3d) = 25 – 15\)
\(4d = 10 \implies d = 2.5\)
Substitute \(d = 2.5\) into (1):
\(a + 3 \cdot 2.5 = 15 \implies a + 7.5 = 15 \implies a = 7.5\)
30th term: \(a + 29d = 7.5 + 29 \cdot 2.5 = 7.5 + 72.5 = 80\)
Answer: \(80\)

Question

Find the term independent of \(x\) in the expansion of each of the following:
(a) \(\left(x + \frac{3}{x^2}\right)^6\)
(b) \((4x^3 – 5)\left(x + \frac{3}{x^2}\right)^6\)

▶️ Answer/Explanation

Solution

(a) For \(\left(x + \frac{3}{x^2}\right)^6\), general term: \(\binom{6}{r} x^{6-r} \left(\frac{3}{x^2}\right)^r = \binom{6}{r} 3^r x^{6-r-2r} = \binom{6}{r} 3^r x^{6-3r}\).
Independent of \(x\): \(6 – 3r = 0 \implies r = 2\).
Term: \(\binom{6}{2} \cdot 3^2 = 15 \cdot 9 = 135\).
Answer (a): \(135\)

(b) For \((4x^3 – 5)\left(x + \frac{3}{x^2}\right)^6\), use the general term from (a): \(\binom{6}{r} 3^r x^{6-3r}\).
Steps:
1. First term: \(4x^3 \cdot \binom{6}{r} 3^r x^{6-3r} = 4 \binom{6}{r} 3^r x^{9-3r}\). Set \(9 – 3r = 0 \implies r = 3\).
Term: \(4 \cdot \binom{6}{3} \cdot 3^3 = 4 \cdot 20 \cdot 27 = 2160\).
2. Second term: \(-5 \cdot \binom{6}{r} 3^r x^{6-3r}\). Set \(6 – 3r = 0 \implies r = 2\).
Term: \(-5 \cdot \binom{6}{2} \cdot 3^2 = -5 \cdot 15 \cdot 9 = -675\).
3. Total: \(2160 – 675 = 1485\).
Answer (b): \(1485\)
Final Answer: \((a) 135; (b) 1485\)

Question

The first term of an arithmetic progression is -20 and the common difference is 5.
(a) Find the sum of the first 20 terms of the progression.
(b) Given that the sum of the first \(2k\) terms is 10 times the sum of the first \(k\) terms, find the value of \(k\).

▶️ Answer/Explanation

Solution

Given: first term \(a = -20\), common difference \(d = 5\).
(a) Sum of first \(n\) terms: \(S_n = \frac{n}{2} (2a + (n-1)d)\).
For \(n = 20\):
\(S_{20} = \frac{20}{2} (2(-20) + (20-1) \cdot 5) = 10 (-40 + 95) = 10 \cdot 55 = 550\).
Answer (a): \(550\)

(b) Given: \(S_{2k} = 10 \cdot S_k\).
Steps:
1. Sum of first \(k\) terms: \(S_k = \frac{k}{2} (2(-20) + (k-1) \cdot 5) = \frac{k}{2} (5k – 45)\).
2. Sum of first \(2k\) terms: \(S_{2k} = \frac{2k}{2} (2(-20) + (2k-1) \cdot 5) = k (10k – 45)\).
3. Set up equation: \(k (10k – 45) = 10 \cdot \frac{k (5k – 45)}{2}\).
4. Divide by \(k\) (since \(k \neq 0\)): \(10k – 45 = 5 (5k – 45)\).
5. Solve: \(10k – 45 = 25k – 225 \implies 180 = 15k \implies k = 12\).
Answer (b): \(12\)
Final Answer: \((a) 550; (b) 12\)

Question

An arithmetic progression has first term 5 and common difference \(d\), where \(d > 0\). The second, fifth, and eleventh terms of the arithmetic progression, in that order, are the first three terms of a geometric progression.
(a) Find the value of \(d\).
(b) The sum of the first 77 terms of the arithmetic progression is denoted by \(S_{77}\). The sum of the first 10 terms of the geometric progression is denoted by \(G_{10}\). Find the value of \(S_{77} – G_{10}\).

▶️ Answer/Explanation

Solution

(a) Arithmetic progression (AP): first term \(a_1 = 5\), common difference \(d > 0\). Terms:
Second term: \(a_2 = 5 + d\)
Fifth term: \(a_5 = 5 + 4d\)
Eleventh term: \(a_{11} = 5 + 10d\)
These form a geometric progression (GP), so: \(\frac{5 + 4d}{5 + d} = \frac{5 + 10d}{5 + 4d}\).
Cross-multiply: \((5 + 4d)^2 = (5 + d)(5 + 10d)\).
Expand: \(25 + 40d + 16d^2 = 25 + 55d + 10d^2\).
Simplify: \(16d^2 + 40d = 10d^2 + 55d \implies 6d^2 – 15d = 0 \implies 3d (2d – 5) = 0\).
Since \(d > 0\), \(d = \frac{5}{2}\).
Answer (a): \(\frac{5}{2}\)

(b) Find \(S_{77} – G_{10}\).
Steps:
1. AP sum (\(S_{77}\)): \(S_n = \frac{n}{2} (a_1 + a_n)\), \(a_1 = 5\), \(d = \frac{5}{2}\).
\(a_{77} = 5 + 76 \cdot \frac{5}{2} = 5 + 190 = 195\).
\(S_{77} = \frac{77}{2} (5 + 195) = \frac{77}{2} \cdot 200 = 7700\).
2. GP sum (\(G_{10}\)): first term \(a_2 = 5 + \frac{5}{2} = \frac{15}{2}\), ratio \(r = \frac{5 + 4 \cdot \frac{5}{2}}{5 + \frac{5}{2}} = \frac{15}{\frac{15}{2}} = 2\), \(n = 10\).
\(G_{10} = \frac{15}{2} \cdot \frac{2^{10} – 1}{2 – 1} = \frac{15}{2} \cdot 1023 = \frac{15345}{2}\).
3. Difference: \(S_{77} – G_{10} = 7700 – \frac{15345}{2} = \frac{15400}{2} – \frac{15345}{2} = \frac{55}{2}\).
Answer (b): \(\frac{55}{2}\)
Final Answer: \((a) \frac{5}{2}; (b) \frac{55}{2}\)

Question

In the expansion of \(\left(kx + \frac{2}{x}\right)^4\), where \(k\) is a positive constant, the term independent of \(x\) is equal to 150. Find the value of \(k\) and hence determine the coefficient of \(x^2\) in the expansion.

▶️ Answer/Explanation

Solution

General term of \(\left(kx + \frac{2}{x}\right)^4\): \(\binom{4}{r} (kx)^{4-r} \left(\frac{2}{x}\right)^r = \binom{4}{r} k^{4-r} 2^r x^{4-2r}\).
Constant term (\(x^0\)):
Set \(4 – 2r = 0 \implies r = 2\).
Term: \(\binom{4}{2} k^{4-2} 2^2 = 6 \cdot k^2 \cdot 4 = 24k^2 = 150\).
Solve: \(k^2 = \frac{150}{24} = \frac{25}{4} \implies k = \frac{5}{2}\) (positive).
Coefficient of \(x^2\):
Set \(4 – 2r = 2 \implies r = 1\).
Term: \(\binom{4}{1} k^{4-1} 2^1 = 4 \cdot k^3 \cdot 2\).
With \(k = \frac{5}{2}\), \(k^3 = \frac{125}{8}\): \(4 \cdot \frac{125}{8} \cdot 2 = \frac{1000}{8} = 125\).
Final Answer: \(k = \frac{5}{2}, \text{ coefficient of } x^2 = 125\)

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