A man saves ? 145000 in ten years. In each year after the first year. he saved ? 2000 more than he did in the proceeding year. How much did he save in the first year?

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 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.

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

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
12 x 2 = 24
24 x 3 = 72
72 x 2 = 144
144 x 3 = 432
432 x 2 = 864
Should come in place of '?'

Here, the 1st AP is (1 + 6 + 11 + ...)
and 2nd AP is (4 + 5 + 6 + ...)

1st AP = (1 + 6 + 11 + ..)
Here , common difference = 5 and the number of terms = 100
? Sum of series = S₁ = n/2[2a + (n - 1)d]
= 100/2 [2 x 1 + (100 - 1) x 5 ]
= 50[2 + 99 x 5 ]
= 50 x 497
= 24850

2nd AP = (4 +5 + 6 + ...)
Here, common difference = 1
and number of terms = 100
S₂ = 100/2[2 x 4 + 99 x 1]
= 50 x 107 = 5350
? Sum of the given series
= S₁ + S₂ = 24850 + 5350
= 30200

1st series is 1, 3, 5, 7 , .... 120 terms
Last term = a + (120 -1) x d = 1 + 119 x 2 = 239
Hence, 1, 3, 5, 7,..., 239 is the first series.

2nd series 3, 6, 9, .., 80 term.
Last term = 3 + (80 - 1) x 3 = 240
Hence, 3, 6 , 9 ....240 is the last term

From above two series, the first common term is 3 and next is 9 and so on, Hence , the series of common terms is 3, 9, 15,.. last term.

Last common term is 237 (occurring in both).
Thus, the number of terms is this series which will give the number of common terms between the two series
(l - a)/(d + 1) = (237 - 3)/(6 + 1) = 40

Put n = 1, now the sum of the series = 1.2.4 = 8
Put n = 1, in the options
(a) 1(1 + 1) (1 + 2) = 6
(b) 1(1 + 1)/12 x (3 + 19 + 26) = 8
(c) (1 + 1) (1 + 2) (1 + 3)/4 = 6
(d) 1(1 + 1) (1 + 2) (1 + 3 )/3 = 8
As, the sum of series = 8
Hence, option (a) and (c) can be rejected.
Now, put n = 2
Sum 1. 2. 4 + 2 . 3 . 5 = 38
Put n = 2, in option (b)
2(2 + 1)/12 (3 x 4 + 19 x 2 + 26)38
Hence, (b) is the correct option.

Let a be the first term and and r be the common ratio of the GP. From the given problem,
a + ar^{n - 1} = 66 ...(i)
Also, ar x ar^{n - 2} = 128 ? a²r^{n - 1} = 128 ... (ii)
From Eq. (ii),
a.ar^{n - 1} = 128
ar^{n - 1} = 128/ a
On substitution this in Eq.(i), we get
a + 128/a = 66
a² - 66a + 128 = 0
a = [-b ± ?b² - 4ac] /2a
= 66 ± ?66² - 4 x 128 x 1)/2 = 64 or 2

The 1st term of an HP is 1/17 .
Hence, for the corresponding AP, the terms is 17 .
Here, a = 17
nth term of an AP is given by T_n = a + (n -1)d
? T₂ = 17 + d, T₄ = 17 + 3d
T₅ = 17 + 4d, T₆ = 17 + 5d
? (17 + d) (17 + 3d) = (17 + 4d) (17 + 5d)
? 289 + 51d + 17d + 3d² = 289 + 85d + 68d + 20d²
? 68d + 3d² = 153d + 20d²
? 17d² = -85d
? d = -5
? 3rd term is T₃ = a + 2d = 17 + 2(-5)
= 17 - 10 = 7
Hence, for the corresponding HP, the 3rd term is 1/7.