Number Series

The series follows the pattern of doubling the previous term (4*2=8, 8*2=16, 16*2=32). The next terms are 32*2=64 and 64*2=128.

The series consists of squares of consecutive integers: 1^2=1, 2^2=4, 3^2=9, 4^2=16. The next two terms are 5^2=25 and 6^2=36.

The series follows the pattern: 3 * 1 = 3, 3 * 2 = 6, 6 * 1.5 = 9, 9 * 1.66... (this is not right). Let's look at differences: 0, 3, 3, 6, 9. The pattern is that each term (starting from the third) is the sum of the two preceding terms: 3+3=6, 3+6=9, 6+9=15, 9+15=24.

The series is powers of 3: 3^1, 3^2, 3^3, 3^4, 3^5, 3^6. Missing terms are 3^4 = 81 and 3^6 = 729.

This is an arithmetic progression with the first term a = 14 and a common difference d = 18 - 14 = 4. The nth term is given by a + (n - 1)d. For the 6th term, it is 14 + (6 - 1) * 4 = 14 + 20 = 34.

This is an arithmetic progression with a = 30 and d = 4. The 27th term = a + (n-1)d = 30 + (26 * 4) = 30 + 104 = 134.

The pattern is: 12+13=25, 13+25=38, 25+38=63, 38+63=101. The series should be 12, 13, 25, 38, 63, 101, 164, 265. The number 104 is incorrect.

This is an arithmetic progression with first term a=5 and common difference d=3. The n-th term is a + (n-1)d. For n=20, term = 5 + 19*3 = 5 + 57 = 62.

Sequence: 4, 6, 10, 16. Differences: 2, 4, 6. This is a quadratic sequence. a_n = an^2 + bn + c. a_1 = a+b+c=4, a_2 = 4a+2b+c=6, a_3 = 9a+3b+c=10. Subtracting: 3a+b=2, 5a+b=4. 2a=2, a=1. b=-1. c=4. a_n = n^2 - n + 4. a_20 = 400 - 20 + 4 = 384.

Term = 2n - 5. First 5 terms: n=1 to 5: -3, -1, 1, 3, 5. Sum = 5. Next 5 terms: n=6 to 10: 7, 9, 11, 13, 15. Sum = 55. Ratio = 5/55 = 1/11.

If S_n = (2n+3)/(6n+5), the ratio of the n-th terms is given by ratio of S_n / S_m where m = 2n-1. For the 13th term, n=13, so we use the ratio (2(25)+3) / (6(25)+5) = 53 / 155.

The pattern is: 11*2-1=21, 21*2-1=41, 41*2-1=81, 81*2-1=161. The missing term is 21.