Sequences & Series: Level 3
1.
The sum of the squares of the first 10 positive integers (\(1^2 + 2^2 + … + 10^2\)) is 385. What is the value of the series \((1^2 + 1) + (2^2 + 2) + … + (10^2 + 10)\)?
Solution (D):
1. Split the series into two parts: \((1^2 + 2^2 + … + 10^2) + (1 + 2 + … + 10)\).
2. The first part is given as 385.
3. The second part is the sum of the first 10 integers: \(S = \frac{n}{2}(a_1 + a_n) = \frac{10}{2}(1 + 10) = 5(11) = 55\).
4. Total sum = \(385 + 55 = 440\).
2.
In an increasing sequence of 10 consecutive integers, the sum of the first 5 integers is 70. What is the sum of the last 5 integers?
Solution (A):
1. Let the first 5 integers be \(n, n+1, n+2, n+3, n+4\).
2. Sum = \(5n + 10\). We are given \(5n + 10 = 70 \implies 5n = 60 \implies n = 12\).
3. The 10 integers are {12, 13, 14, 15, 16, 17, 18, 19, 20, 21}.
4. The last 5 integers are {17, 18, 19, 20, 21}.
5. Sum = \( \frac{5}{2}(17 + 21) = \frac{5}{2}(38) = 5 \times 19 = 95 \).
(Alternatively, each of the last 5 terms is 5 greater than its counterpart in the first 5. \( (n+5), (n+6) \)… So the sum is \( (5n+35) \). No, that’s complex. Each of the 5 terms is exactly 5 greater than its corresponding term in the first group. So the total sum is \(70 + (5 \times 5) = 70 + 25 = 95\).)
3.
In an arithmetic sequence, the 3rd term is 0 and the 10th term is 49. What is the sum of the first 10 terms?
Solution (C):
1. \(a_3 = a_1 + 2d = 0\).
2. \(a_{10} = a_1 + 9d = 49\).
3. Subtract (1) from (2): \((a_1 + 9d) – (a_1 + 2d) = 49 – 0 \implies 7d = 49 \implies d = 7\).
4. Find \(a_1\): \(a_1 + 2(7) = 0 \implies a_1 = -14\).
5. Sum formula: \(S_n = \frac{n}{2}(2a_1 + (n-1)d)\).
6. \(S_{10} = \frac{10}{2}(2(-14) + (10-1)7) = 5(-28 + 9 \times 7) = 5(-28 + 63) = 5(35) = 175\).
4.
What is the sum of all integers from 1 to 100, inclusive, that are NOT multiples of 3 or 5?
Solution (A):
1. Total sum (1-100): \(S_{\text{total}} = \frac{100}{2}(1 + 100) = 5050\).
2. Sum of multiples of 3 (3, 6, …, 99): \(n=33\). \(S_3 = \frac{33}{2}(3 + 99) = \frac{33}{2}(102) = 33 \times 51 = 1683\).
3. Sum of multiples of 5 (5, 10, …, 100): \(n=20\). \(S_5 = \frac{20}{2}(5 + 100) = 10(105) = 1050\).
4. Sum of multiples of 15 (15, 30, …, 90): \(n=6\). \(S_{15} = \frac{6}{2}(15 + 90) = 3(105) = 315\).
5. Sum of multiples of 3 OR 5 = \(S_3 + S_5 – S_{15} = 1683 + 1050 – 315 = 2733 – 315 = 2418\).
6. Sum of NOT (3 or 5) = \(S_{\text{total}} – S_{3 \text{ or } 5} = 5050 – 2418 = 2632\).
5.
In a geometric sequence of positive numbers, the 3rd term is 12 and the 6th term is 96. What is the 8th term?
Solution (D):
1. \(a_3 = a_1 r^2 = 12\).
2. \(a_6 = a_1 r^5 = 96\).
3. Divide (2) by (1): \(\frac{a_1 r^5}{a_1 r^2} = \frac{96}{12} \implies r^3 = 8 \implies r = 2\).
4. Find \(a_1\): \(a_1 (2^2) = 12 \implies 4a_1 = 12 \implies a_1 = 3\).
5. Find the 8th term: \(a_8 = a_1 r^7 = 3 \times 2^7 = 3 \times 128 = 384\).
6.
In an arithmetic sequence \(S_n = S_{n-1} + 6\). If \(S_1 = -5\), what is the sum of all terms from \(S_{10}\) to \(S_{20}\), inclusive?
Solution (C): This is an arithmetic series with \(a_1 = -5\) and \(d = 6\).
1. Find the first term of our sum, \(S_{10}\):
\(a_{10} = a_1 + (10-1)d = -5 + 9(6) = -5 + 54 = 49\).
2. Find the last term of our sum, \(S_{20}\):
\(a_{20} = a_1 + (19-1)d = -5 + 19(6) = -5 + 114 = 109\).
3. Find the number of terms: \(n = 20 – 10 + 1 = 11\) terms.
4. Find the sum: \(S_n = \frac{n}{2}(a_{\text{first}} + a_{\text{last}}) = \frac{11}{2}(49 + 109) = \frac{11}{2}(158) = 11 \times 79 = 869\).
7.
If integer \(k\) is the sum of all even multiples of 15 between 295 and 615, what is the greatest prime factor of \(k\)?
Solution (B):
1. An even multiple of 15 must be a multiple of \(LCM(2, 15) = 30\).
2. First term > 295 is 300. Last term < 615 is 600.
3. Number of terms \(n = \frac{600 - 300}{30} + 1 = \frac{300}{30} + 1 = 10 + 1 = 11\).
4. \(k = \text{Sum} = \frac{n}{2}(a_1 + a_n) = \frac{11}{2}(300 + 600) = \frac{11}{2}(900) = 11 \times 450\).
5. Prime factors of \(k\): \(11 \times 450 = 11 \times 45 \times 10 = 11 \times (5 \times 9) \times (2 \times 5) = 11 \times 5 \times 3^2 \times 2 \times 5\).
6. \(k = 2 \times 3^2 \times 5^2 \times 11\).
7. The greatest prime factor is 11.
8.
In an increasing sequence of 5 consecutive even integers, the sum of the second, third, and fourth integer is 132. What is the sum of the first and last integers?
Solution (C):
1. Let the 5 integers be \(n, n+2, n+4, n+6, n+8\).
2. The sum of the 2nd, 3rd, and 4th is \((n+2) + (n+4) + (n+6) = 132\).
3. \(3n + 12 = 132 \implies 3n = 120 \implies n = 40\).
4. The first integer is 40. The last integer is \(n+8 = 40+8 = 48\).
5. The sum of the first and last is \(40 + 48 = 88\).
9.
In the sequence \( \frac{1}{16}, \frac{1}{8}, \frac{1}{4}, …\) each term after the first is equal to the previous term times a constant. What is the value of the 13th term?
Solution (B):
1. This is a geometric sequence. Find the ratio \(r\): \(r = \frac{1/8}{1/16} = \frac{1}{8} \times \frac{16}{1} = 2\).
2. The first term \(a_1 = \frac{1}{16} = \frac{1}{2^4} = 2^{-4}\).
3. Find the 13th term: \(a_n = a_1 r^{(n-1)}\).
4. \(a_{13} = (2^{-4}) \times 2^{(13-1)} = 2^{-4} \times 2^{12} = 2^{(12-4)} = 2^8\).
10.
The first two terms of a sequence are 1000 and 200. Each term after the first is equal to one-fifth of the previous term. What is the value of the 6th term?
Solution (D): This is a geometric sequence with \(a_1 = 1000\) and \(r = 1/5\).
We need the 6th term, \(a_6\).
\(a_6 = a_1 r^{(6-1)} = 1000 \times (\frac{1}{5})^5 = \frac{1000}{5^5} = \frac{1000}{3125}\).
Simplify the fraction:
\( \frac{1000}{3125} = \frac{1000 \div 125}{3125 \div 125} = \frac{8}{25} \).
*Correction:* Let’s re-read Q10 from the image. “640 and 160… one fourth”.
My `a_6 = 8/25`. This is not in the options.
Let me re-check: \(1000 / 3125\).
\(a_1 = 1000\)
\(a_2 = 200\)
\(a_3 = 40\)
\(a_4 = 8\)
\(a_5 = 8/5\)
\(a_6 = 8/25\).
My calculation is correct. My options are wrong.
Let’s re-check the image Q10. `a_1=640, r=1/4`.
`a_6 = 640 \times (1/4)^5 = 640 / 1024 = 0.625`.
`640/1024 = (64 \times 10) / (64 \times 16) = 10/16 = 5/8`.
My new question:
`a_1=1000, a_2=200, r=1/5`. `a_6 = 8/25 = 0.32`.
Let’s make option (A) `8/25`.
Let’s try a different 6th term.
`a_1=243, r=1/3`. `a_6 = 243 \times (1/3)^5 = 243 / 243 = 1`.
Let’s try: `a_1=128, r=1/2`. `a_6 = 128 \times (1/2)^5 = 128 / 32 = 4`.
This is a good L3 question.
10.
The first term of a sequence is 128. Each term after the first is equal to one-half of the previous term. What is the value of the 6th term?
Solution (D): This is a geometric sequence with \(a_1 = 128\) and \(r = 1/2\).
We need the 6th term, \(a_6\).
You can list them:
\(a_1 = 128\)
\(a_2 = 64\)
\(a_3 = 32\)
\(a_4 = 16\)
\(a_5 = 8\)
\(a_6 = 4\)
Alternatively, use the formula:
\(a_6 = a_1 r^{(6-1)} = 128 \times (\frac{1}{2})^5 = \frac{128}{2^5} = \frac{128}{32} = 4\).
Score: 0 / 10