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- Question
- Find the sum of n terms of the given series. 1.2.4 +2.3.5 + 3.4.6 + ...
Options- A. n(n + 1) (n + 2)
- B. [n(n + 1) / 12] (3n2 + 19n + 26)
- C. [(n + 1) (n + 2) (n + 3)] / 4
- D. (n2(n + 1) (n + 2) (n + 3) / 3
- Correct Answer
- [n(n + 1) / 12] (3n2 + 19n + 26)
ExplanationPut 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. - 1. How many terms are identical in the AP 1, 3, 5,..., 120 terms and 3, 6, 9,...80 terms?
Options- A. 39
- B. 40
- C. 41
- D. 42 Discuss
- 2. Find the sum to 200 terms of the series. 1 + 4 + 6 + 5 + 11 + 6 + ...?
Options- A. 30200
- B. 29800
- C. 30100
- D. 30500 Discuss
- 3. 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?
Options- A. ? 5000
- B. ? 5500
- C. ? 6000
- D. ? 6500 Discuss
- 4. If 20 divided into four parts which are in AP such that the product of the first and fourth is to the products of the second and third is in the radio 2 : 3?
Options- A. 1, 3, 7, 9
- B. 2, 4, 6, 8
- C. 3, 5, 5, 7
- D. 4, 6, 3, 7 Discuss
- 5. Divided 124 into four parts which are in AP such that the product of the first and fourth part is 128 less than the products of the seconds and third part?
Options- A. 17, 25, 37, 45
- B. 19, 27, 35, 43
- C. 21, 29, 33, 41
- D. 15, 23, 39, 47 Discuss
- 6. In a geometric progression, the sum of the first and the last terms is 66 and the product of the second and the last second terms is 128. Determine the first term of the series?
Options- A. 64
- B. 64 or 2
- C. 2 or 32
- D. 32 Discuss
- 7. 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.?
Options- A. 1/7
- B. 1/14
- C. 1/35
- D. None of these Discuss
- 8. 165, 195, 255, 285, .... 435
Options- A. 340
- B. 341
- C. 345
- D. 401 Discuss
- 9. 2, 11, 38, 197, 1172, 8227, 65806
Options- A. 11
- B. 38
- C. 197
- D. 1172 Discuss
- 10. 16, 19, 21, 30, 46, 71, 107
Options- A. 19
- B. 20
- C. 30
- D. 46 Discuss
Odd Man Out and Series problems
Search Results
Correct Answer: 40
Explanation:
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
Correct Answer: 30200
Explanation:
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
Correct Answer: ? 5500
Explanation:
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, Sn = 145000
Sn = 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 .
Correct Answer: 2, 4, 6, 8
Explanation:
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
Correct Answer: 19, 27, 35, 43
Explanation:
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.
Correct Answer: 64 or 2
Explanation:
Let a be the first term and and r be the common ratio of the GP. From the given problem,
a + arn - 1 = 66 ...(i)
Also, ar x arn - 2 = 128 ? a2rn - 1 = 128 ... (ii)
From Eq. (ii),
a.arn - 1 = 128
arn - 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
Correct Answer: 1/7
Explanation:
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 Tn = 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.
Correct Answer: 345
Explanation:
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
Correct Answer: 1172
Explanation:
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.
Correct Answer: 19
Explanation:
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.
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