Find the missing number in this series 6, 12, 18, 49, 169, 361, 512, ?, 8000

M4maths Previous Puzzles

31 Oct 2016 Sequence

Find the missing number in this series
6, 12, 18, 49, 169, 361, 512, ?, 8000

OPtion

1) 5832
2) 729
3) 1000
4) 2197
5) 3375
6) 2744
7) 4913
8) 1331
9) 4095
10)6843

Solution

Reason
Series = (x,2x,3x),((x+1)^2,(2x+1)^2,(3x+1)^2),((x+2)^3,(2x+2)^3,(3x+2)^3)
Clearly x=6
Therefore the missing number = 2744

Option 6)

Correct Option: 6 Best Solution (22)

G.S.Kantharao   8 years AGO

(6,12,18),(49,169,361),(512,?,8000)
In the first set (6,12,18) divisible by 6 leaves 0 remainder.
In the second set (49,169,361) divisible by 6 leaves 1 as remainder.
In the third set ( 512,?,8000) which should be ( 512,2744,8000) divisible by 6 leaves 2 as remainder.
Hence my option is 6.

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Mukund Manohar   8 years AGO

6(6¹) 12(12¹) 18(18¹)
49(7²) 169(13²) 361(19²)
512(8³) 2744(14³) 8000(20³)
2744 is the missing number

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dipanwita Bishnu   8 years AGO

6, 12,18,
(6+1)^2=49
(12 +1)^2=169
(18+1)^2= 361
(6+2)^3= 512
So
(12+2)^3= 2744

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viswambher   8 years AGO

1st three numbers 6 12 18
2nd three numbers 7^2 13^2 19^2
3rd three numbers 8^3 14^3 20^3
14^3 = 2744

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Dinesh kumar sah   8 years AGO

Simply,
(6,12,18),(49,169,361),(512,?,8000)
make group of first,2nd n third of each group. ie
(6,49,512)(12,169,?)(18,361,8000)
so in ? must 14^3 ie 2744 Ans
Hence option(6) is the right answer.

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YASWANTH REDDY   8 years AGO

the series can be written as
6,12,18,
(6+1)^2, (12+1)^2, (18+1)^2,
(6+2)^3,______, (18+2)^3.
following the same pattern, the missing number will be (12+2)^3 = 14^3= 2744

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Ferman Naqvi   8 years AGO

6 1=7 and 7^2=49
12 1=13 and 13^2=169
18 1=19 and 19^2=361
6 2=8 and 8^3=512
12 2=14 and 14^3=(2744)
18 2=20 and 20^3=8000

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Amit kumar singh   8 years AGO

6^1,12^1,18^1,7^2,13^2,19^2,8^3,?,20^3
So missing term=14^3=2744

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Nitish kesarwani   8 years AGO

taking each (3 pairs of series ) on dividing by 6
leaving reminder
(0 0 0) (1 1 1 ) (2 2 2)
so option 6

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Nandini JB   8 years AGO

(6,12,18) ,(49,169,361) , (512,?,8000)
6^1=6, 12^1=12, 18^1=18
7^2=49, 13^2=169, 19^2=361
8^3=512, 14^3=2744, 20^3=8000

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Dinesh Sharma   8 years AGO

6^1,12^1,18^1
7^2,13^2,19^2
8^3,14^3,20^3

So, Missing value = 14^3 = 2744. Option(6)

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sourabh nath agrawal   8 years AGO

Bcoz every has 6 increasing in 4th pace to 6th pace every time 6 incresing with square same in 7th to 9th 6 increse with cube

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Hemanth   8 years AGO

6, 12, 18 ............ Difference b/w numbers is 6
7*7, 13*13, 19*19..........
8*8*8, 14*14*14 (2744), 20*20*20

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Samjukta Das   8 years AGO

6+1=7*7=49
12+1=13=13*13=169
18+1=19=19*19=361
6+2=8=8*8*8=512
12+2=14=14*14*14=2744
18+2=20=20*20*20=8000

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SATHEESH V   8 years AGO

(6,12,18),(49,169,361),(512,?,8000)

(6+1)^2=49
(12+1)^2=169
(18+1)^2=361

(6+2)^3=512
(12+2)^3=2744
(18+2)^3=8000
therefore,
answer =2744

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deepak agrawal   8 years AGO

6^1=6
(6+6)^1=12
(12+6)^1=18
7^2=49
(7+6)^2=169
(13+6)^2=361
8^3=512
(8+6)^3=2744
(14+6)^3=8000

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RAM NATH YADAV   8 years AGO

The series is of type
6, 6+6, 6+6+6, (7)^2, (7+6)^2, (7+6+6)^2, (8)^3, (8+6)^3, (8+6+6)^3

6, 12, 18, 49, 169, 361, 512, 2744, 8000

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VED PRAKASH SINGH   8 years AGO

6+1=7^2=49
12+1=13^2=169
18+1=19^2=361
6+2=8^3=512
12+2=14^3=2744

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Jayendra   8 years AGO

6*1
6*2
6*3
7^2
13^2
19^2
8^3
14^3
20^3

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ANAND MOHAN RAI   8 years AGO

6,12,18 with exponent 1 and difference..then 7^2,13^2,19^2, i.e exponent 2 and difference 6,then 8^3,14^3,20^3....hence 14^3=2744
option 6

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Sourav Das   8 years AGO

first three no. are randomly chosen.4th no. is (1st no.+1)^2 .that is (6+1)^2 =49..so are 5th & 6th no. 7th is (1st no.+2)^3 that is (6+2)^3 that is 512 ..so the 8th & 9th no.

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shanthi   8 years AGO

6
6*2=12
6*3=18

49=7^2
169=13^2
361=19^2
7-13=13-19=>6

512=8^3
8000=20^3
8-20=12=>12/2=6
thus,8+6=14=>so,14^3=2744

so,ans is 2744

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