12, 24, 72, 144, 432,?,

Series pattern
600/5 + 5 = 125
125/5 + 5 = 30
30/5 + 5 = 11
Should come in place of '?'
11/5 + 5 = 7.2
7.5/5 + 5 = 6.44
6.44/5 + 5 = 6.288

1 x 1 + 1 = 2
2 x 2 + 2 = 6
6 x 3 + 3 = 21
21 x 4 + 4 = 88
88 x 5 + 5 = 445
445 x 6 + 6 = 2676 Should come in place of '?'

Series pattern
50 x 1.2 = 60,
60 x 1.25 = 75,
75 x 1.3 = 97.5,
97.5 x 1.35 = 131.625 Should come in place of '?'
131.625 x 1.4 = 184.275
184.275 x 1.45 = 267.19875

Series pattern
1³ + 5, 2³ + 5, 3³ + 5, 4³ + 5, 5³ + 5
? Missing term = 4³ + 5 = 69

Series pattern
656 - 224 = 432
432 - 112 = 320
320 - 56 = 264
264 - 28 = 236
236 - 14 = 222 Should come in place of '?'

Series pattern
39 + 1 x 13 = 52
52 + 2 x 13 = 78
78 + 3 x 13 = 117
117 + 4 x 13 = 169
169 + 5 x 13 = 234 Should come in place of ?

Series Pattern
(-1)², (-2)², (-3)², (-4)²
Missing term = 36 - 3² = 27

Let the four parts be (a - 3d), (a - d), (a + d) and (a + 3d).
(a - 3d) + (a - d) + (a +d ) + (a + 3d) = 124
? a = 31
Also, (a - 3d) (a + 3d) = (a - d) (a + d) - 128
? a² - 9d² = a² - d² - 128
? 8d² = 128
? d = 4
a = 31, d = 4
So, the four parts are 19, 27, 35, 43.

Let the four parts be (a - 3d), (a - d), (a +d), (a + 3d).
From question,
(a -3d) + (a - d) + (a + b) + (a + 3d) = 20
? a = 5
Also, [(a - 3d) (a +3d)] / [(a - d) (a + d)] = 2/3
? [a² - 9d²] / [a² - d²] = 2/3
? [5² - 9d²] / [5² - d²] = 2/3
? d = 1
? a = 5, d = 1
So, the four parts are 2, 4, 6, 8

Let the man first save ? P in the first year.
Then, the given sequence is P + (P + 2000) + (P + 4000) + ..,
which is an AP with a = P , d = (P + 2000) - P = 2000,
n = 10, S_n = 145000

S_n = n/2[2a + (n - 1)d]
? 145000 = 10/2 [2 x P + 9 x 2000] = 5 [2P + 18000]
? 2P + 18000 = 145000/5 = 29000
? 2P = 29000 - 18000
? 2P = 11000
? P = ? 5500
So, the man save ? 5500 in the first year .