Number Series

Series: 3, 5, 12, 38, 154, 914, 4634. Pattern: *1+2=5, *2+2=12, *3+2=38, *4+2=154, *5+2=772 (not 914), *6+2=4634. 914 is the incorrect term.

The differences between consecutive terms in the series should be consecutive perfect squares: +1^2, +2^2, +3^2, +4^2, and so on. Adding 9 (which is 3^2) to 144 should give 153 instead of 156, which then correctly leads to 153 + 16 = 169.

This is an arithmetic sequence with first term a = 4 and common difference d = 3. The nth term is a + (n-1)d = 4 + (n-1)3 = 4 + 3n - 3 = 3n + 1.

The series consists of two alternating sequences: 2, 7, 12, ... (adding 5) and 65, 59, 53, ... (subtracting 6). The next terms are 12+5=17 and 53-6=47.

The series increases by 51 each time: 5255+51=5306, 5306+51=5357, 5357+51=5408, 5408+51=5459.

In an arithmetic sequence, the difference between consecutive terms is constant. 2a - a = a. So the common difference is a. Third term b = 2a + a = 3a. Fourth term = 3a + a = 4a. Given 4a = a - 6 - b. Substituting b = 3a: 4a = a - 6 - 3a => 4a = -2a - 6 => 6a = -6 => a = -1. Sequence: -1, -2, -3, -4. 100th term = a + (n-1)d = -1 + 99(-1) = -100.

The n-th term of an AP is a + (n-1)d. For the 4th term, T4 = a + 3d. The difference between the 4th terms is (a1 + 3d) - (a2 + 3d) = a1 - a2. Given a1 = -1 and a2 = -8, the difference is -1 - (-8) = 7.

The series is an AP with a=3 and d=3. The sum formula is S = n/2 * (2a + (n-1)d). Setting 108 = n/2 * (6 + (n-1)3) leads to 108 = n/2 * (3n + 3), which simplifies to 3n^2 + 3n - 216 = 0, or n^2 + n - 72 = 0. Factoring gives (n+9)(n-8) = 0, so n=8.

The pattern is (x - 3) / 2 = next term. 445 -> 221 (445-3)/2 = 221. 221 -> 109 (221-3)/2 = 109. 109 -> 53 (109-3)/2 = 53. 53 -> 25 (53-3)/2 = 25. 25 -> 11 (25-3)/2 = 11. 11 -> (11-3)/2 = 4.

The pattern is: 12*2+2=26, 26*2+4=56, 56*2+6=118, 118*2+8=244, 244*2+10=498, 498*2+12=1008.

Sum = n/2 * (a + l). 114 = n/2 * (37 - 18) = n/2 * 19. 114 = 9.5n. n = 114 / 9.5 = 12.