The 1st term of an HP is 1/17 and the product of the 2nd and 4th term equals to the products of 5th and 6th term of the HP. Find the 3rd term of the HP.?

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

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.

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

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

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 .

Series Pattern
(15 x 11), (15 x 13), (15 x 17), (15 x 19), (15 x 23), (15 x 29)
? Missing term = (15 x 23) = 345

Series Pattern
2 x 3 + 5 = 11, 11 x 4 - 6 = 38
38 x 5 + 7 = 197, 197 x 6 - 8 = 1174
Should come in place of 1172
1174 x 7 + 9 = 8827
8827 x 8 - 10 = 65806
Clearly, 1172 is wrong and will be replaced by 1174.

Series pattern
16 + 1² = 17
Should come in place of 19
17 + 2² = 21 ; 21+ 3² = 30;
30 + 4² = 46 ; 46 + 5² = 71
71 + 6² = 107
Clearly, 19 is wrong and will be replaced by 17.

Series pattern 4 x 0.5 = 2
2 x 1.5 = 3
Should come in place of 3.5
3 x 2.5 = 7.5
7.5 x 3.5 = 26.25
26.25 x 4.5 = 118.125
Clearly, 3.5 is wrong and is replaced by 3.

Series pattern
20 + 1² = 21,
21 + 2² = 25,
25 + 3² = 34,
34 + 4² = 50,
50 + 5² = 75 Should come in place of ' ? '
75 + 6² = 111