Question

1)Find out the wrong number in the series
17 141 358 701 1213 1942
A)141
B)358
C)701
D)1213

2) find out the wrong number
6 14 51 249 1486 10401
A)14
B)51
C)249
D)1486

Answer 1

1.  The given series:

17, 141, 358, 701, 1213, 1942

We have to find, the value of wrong number.

Solution:

The pattern follow:

+ $5^{3}$, + $6^{3}$, + $7^{3}$, + $8^{3}$, + $9^{3}$, ........

141 = 17 + $5^{3}$ = 17 + 125

     = 142 ≠ 141, is wrong number

358 = 142 + $6^{3}$ = 142 + 216

      = 358

701 = 358 + $7^{3}$ = 358 + 343

      = 701

1213 = 701 + $8^{3}$ = 701 + 512

      = 1213

1942 = 1213 + $9^{3}$ = 1213 + 729

        = 1942

∴ Wrong number = 141

Thus, the required option is "A) 141".

2. The given series:

6, 14, 51, 249, 1486, 10401

We have to find, the value of wrong number.

Solution:

The pattern follow:

× 3 - 4, × 4 - 5, × 5 - 6, × 6 - 7, × 7 - 8, ........

14 = 6 × 3 - 4

   = 18 - 4 = 14

51 = 14 × 4 - 5

    = 56 - 5 = 51

249 = 51 × 5 - 6

       = 255 - 6 = 249

1486 = 249 × 6 - 7

        = 1494 - 7 = 1487 ≠ 1486, is wrong number.

10401 = 1487 × 7 - 8

          = 10409 - 8 = 10401

∴ Wrong number = 1486

Thus, the required option is "C) 1486".

Answer 2

Given : series

17 141 358 701 1213 1942

To Find : wrong number in the series

A)141

B)358

C)701

D)1213

Solution:

17    141    358       701    1213      1942

 124     217    343      512      729

5³-1     6³+1     7³        8³        9³

if 5³ - 1  and 6³ + 1  are  5³  and 6³   then it will be a sequence

Tₙ  = Tₙ₋₁  + (n + 3)³

for that 141 should be 142

17    142    358       701    1213      1942

 125     216    343      512      729

5³         6³      7³        8³        9³

141 is wrong Number

6 14 51 249 1486 10401

6 x 3  - 4 =  14

14 x 4 - 5  = 51

51 x 5 - 6  = 249

249 x 6 - 7 = 1487

1487 x 7 - 8 = 10401

1486 is wrong number

sequence is Tₙ = Tₙ₋₁ x(n+1) - (n+ 2)

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