Find the missing number of the series 8, 78, 318, 720, ?, 2010,

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17 Oct 2016 Sequence

Find the missing number of the series
8, 78, 318, 720, ?, 2010,

OPtion

1) 1258
2) 1064
3) 1166
4) 1284
5) 1664
6) 1764
7) 1568
8) 1898
9) 1676
10)1965

Solution

Reason
x = 3
series =>
(1x)^2-1, (3x)^2-3, (6x)^2-6, (9x)^2-9, (12x)^2-12, (15x)^2-15

8, 78, 318, 720, 1284, 2010
Option 4)

Correct Option: 4 Best Solution (8)

lakshmi nalluri   8 years AGO

3^2-1,9^2-3,18^2-6,27^2-9,36^2-12,45^2-15
so ,ans is 36^2-12=1284

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Nwobi Chibuzor    8 years AGO

The format is (3*y)^2 - y.
Where the series y are the consecutive multiples of 3. So y = 1, 3, 6, 9, 12, 15 e.t.c.

(3×1)^2 - 1 = 8
(3×3)^2 - 3 = 78
(3×6)^2 - 6 = 318
(3×9)^2 - 9 = 720
(3×12)^2 - 12 = 1284
(3×15)^2 - 15 = 2010

Therefore, the missing number in the series is 1284.

Like? Yes (2) | No (1)  

Rajesh Durisala   8 years AGO

8 = (3*1)^2 -1
78=(3*3)^2-3
318=(3*6)^2-6
720=(3*9)^2-9
1284=(3*12)^2-12
2010=(3*15)^2-15

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Sagar Kumar   8 years AGO

Ans. is 1284 (4) ... I couldn't find the proper relation between 8 and 78 but from 78 onward i find a relation as explained below...

8= 3^2 -1
78= 9^2 -3
318= 18^2 -6
720=27^2 - 9

So, 36^2 - 12 =1284

at last
2010= 45^2 -15

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Siddartha Mulumudi   8 years AGO

(3)^2 -1 = 8
(3x3)^2-3 = 78
(3*6)^2-6=318
(3*9)^2-9=720
(3*12)^2-12 =1284
(3*15)^2-15=2010

Ans: 1284

Like? Yes | No (1)  

ANAND MOHAN RAI   8 years AGO

the series is based on X^2-(X/3)..
3^2-(3/3)=8
18^2-(18/3)=318
27^2-(27/3)=720
36^2-(36/3)=1284
45^2-(45/3)=2010

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SADDAM   8 years AGO

8
78 》9^2-3
318》(9*2)^2-3*2
720》(9*3) ^2-3*3
?》(9*4)^2-3*4=1284
2010》(9*5)^2-3*5

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Dinesh Sharma   8 years AGO

Constant difference from second element.

i.e. 318-78=240 diff btwn 402 and 240 = 162
720-318=402 so, diff betwn y & 402 = 162
x - 720 = y so, y = 564

i.e x = 1284 and next element 564+162 = 726 ahead from 1284 i.e. 1284+726=2010

Like? Yes | No (4)