Sequences & Series: Level 2
1.
The sum of the squares of the first 10 positive integers (\(1^2 + … + 10^2\)) is 385. What is the sum of the squares of the first 10 positive even integers (\(2^2 + 4^2 + … + 20^2\))?
Solution (D):
The sum is \(S = 2^2 + 4^2 + … + 20^2\).
This can be rewritten as \((2 \cdot 1)^2 + (2 \cdot 2)^2 + … + (2 \cdot 10)^2\).
Factor out the \(2^2\): \(S = 2^2 (1^2 + 2^2 + … + 10^2)\).
We are given that \(1^2 + … + 10^2 = 385\).
\(S = 4 \times (385) = 1540\).
2.
The 5th term of an arithmetic sequence is 18 and the 10th term is 48. What is the 3rd term?
Solution (A):
1. The difference between the 10th and 5th term is \(10 – 5 = 5\) terms (i.e., \(5d\)).
2. The difference in value is \(48 – 18 = 30\).
3. So, \(5d = 30 \implies d = 6\).
4. The 3rd term (\(a_3\)) is 2 terms before the 5th term (\(a_5\)).
5. \(a_3 = a_5 – 2d = 18 – 2(6) = 18 – 12 = 6\).
3.
What is the sum of all the multiples of 3 between 100 and 200 (exclusive of 100 and 200)?
Solution (D):
1. First term (\(a_1\)): Smallest multiple of 3 > 100 is 102.
2. Last term (\(a_n\)): Largest multiple of 3 < 200 is 198.
3. Number of terms (\(n\)): \(\frac{198 - 102}{3} + 1 = \frac{96}{3} + 1 = 32 + 1 = 33\).
4. Sum: \(S_n = \frac{n}{2}(a_1 + a_n) = \frac{33}{2}(102 + 198) = \frac{33}{2}(300) = 33 \times 150 = 4950\).
4.
The first term of a geometric sequence is 5. The 5th term is 80. What is the 3rd term?
Solution (A):
1. Formula: \(a_n = a_1 \times r^{(n-1)}\).
2. We have \(a_1 = 5\) and \(a_5 = 80\).
3. \(a_5 = a_1 \times r^4 \implies 80 = 5 \times r^4 \implies 16 = r^4\).
4. Since the terms are positive, the common ratio \(r = 2\).
5. We need the 3rd term, \(a_3\).
6. \(a_3 = a_1 \times r^2 = 5 \times (2^2) = 5 \times 4 = 20\).
5.
What is the sum of the series \(2^0 + 2^1 + 2^2 + … + 2^7\)?
Solution (E): This is a geometric series with 8 terms (\(2^0\) is the 1st term, \(2^7\) is the 8th).
The sum of a geometric series is \(S_n = a_1 \frac{(r^n – 1)}{(r-1)}\).
Here, \(a_1 = 2^0 = 1\), \(r = 2\), and \(n = 8\).
\(S_8 = 1 \times \frac{(2^8 – 1)}{(2-1)} = 2^8 – 1 = 256 – 1 = 255\).
6.
The sum of 7 consecutive odd integers is 147. What is the product of the smallest and largest integer?
Solution (D):
1. The average of consecutive integers is the middle term.
2. Average = \(147 \div 7 = 21\).
3. Since there are 7 terms, the 4th term (the middle one) is 21.
4. The 7 integers are: {15, 17, 19, 21, 23, 25, 27}.
5. Smallest = 15. Largest = 27.
6. Product = \(15 \times 27 = 405\).
7.
What is the sum of all integers from 1 to 100, inclusive, that are NOT multiples of 4?
Solution (A):
1. Find the total sum (1 to 100): \(S_{\text{total}} = \frac{100}{2}(1 + 100) = 50(101) = 5050\).
2. Find the sum of multiples of 4: The series is 4, 8, …, 100.
3. Number of terms \(n = \frac{100 – 4}{4} + 1 = \frac{96}{4} + 1 = 24 + 1 = 25\).
4. Sum of 4s: \(S_4 = \frac{25}{2}(4 + 100) = \frac{25}{2}(104) = 25 \times 52 = 1300\).
5. Subtract: \(S_{\text{total}} – S_4 = 5050 – 1300 = 3750\).
8.
In a sequence, \(S_1 = 5\) and \(S_n = S_{n-1} + 10\) for \(n > 1\). What is the value of \(S_{100}\)?
Solution (B): This is an arithmetic sequence with first term \(a_1 = 5\) and common difference \(d = 10\). We need to find the 100th term.
Formula: \(a_n = a_1 + (n-1)d\).
\(a_{100} = 5 + (100-1) \times 10 = 5 + (99 \times 10) = 5 + 990 = 995\).
9.
In a sequence, \(S_1 = 10\), \(S_2 = 20\), and \(S_n = S_{n-1} + S_{n-2}\) for \(n > 2\). What is the value of \(S_5\)?
Solution (C): This is a recursive sequence (like Fibonacci).
\(S_1 = 10\)
\(S_2 = 20\)
\(S_3 = S_2 + S_1 = 20 + 10 = 30\)
\(S_4 = S_3 + S_2 = 30 + 20 = 50\)
\(S_5 = S_4 + S_3 = 50 + 30 = 80\)
10.
A certain series is defined by \(S_n = k(S_{n-1})\). If the 1st term is 64 and the 25th term is 192, what is the 9th term?
Solution (A): This is a geometric sequence: \(S_n = a_1 \times k^{(n-1)}\).
We are given \(S_1 = 64\) and \(S_{25} = 192\).
\(S_{25} = S_1 \times k^{(25-1)} \implies 192 = 64 \times k^{24}\).
Divide by 64: \(3 = k^{24}\).
We need to find the 9th term, \(S_9\).
\(S_9 = S_1 \times k^{(9-1)} = 64 \times k^8\).
To find \(k^8\), we manipulate \(k^{24} = 3\).
\(k^8 = (k^{24})^{\frac{1}{3}} = 3^{\frac{1}{3}} = \sqrt[3]{3}\).
Therefore, \(S_9 = 64\sqrt[3]{3}\).
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