▶️ 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: \( 20, \frac{20}{3} \)
▶️ 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: \((a) -90a^3, 15a^4; (b) 2\)
▶️ 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: \(80\)
▶️ 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: \((a) 135; (b) 1485\)
▶️ 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: \((a) 550; (b) 12\)
▶️ 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: \((a) \frac{5}{2}; (b) \frac{55}{2}\)
▶️ 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: \(k = \frac{5}{2}, \text{ coefficient of } x^2 = 125\)