Interleaved series almost always come up in the IPAT and are one of the easiest to recognize and solve. Lets try to explain this series type with an example:

1, 2, 5, 3, 9, 4, ?

If we split this series into two series –

1, 5, 9 (consists of the 1^{st} 3^{rd} and 5^{th} numbers – all odd numbered terms)

2, 3, 4 (consists of the 2^{nd} 4^{th} and 6^{th} numbers –all even numbered terms)

We see a simple difference based pattern in both series. The first series has a constant difference of 4 while the second series has a constant difference of 1. The next number in the parent series (which is the 7^{th} term- an odd number) is the next number in the 1^{st} sub series (1,5,9) and hence is equal to 9 + 4 = 13.

**The rule of 1, 2, 3**

The rule of 1,2,3 gives us a simple framework to approach this type of series. The rule goes as follows: form subseries in the series by first leaving 1 term in between, then 2 terms in between and then 3 terms in between to see which interleaved difference gives a legitimate difference series.

For example, the series above had a subseries gap of 1 term. Here is an example of 2 term subseries

1, 7, 11, 2, 9, 14, 3, 11, 17

Breaking the series into subseries based on 2 term difference

1, 2, 3 (1st, 4th, 7th terms : simple difference series with constant difference of 1)

7, 9, 11 (2nd, 5th, 8th terms : simple difference series with constant difference of 2)

11, 14, 17 (3rd, 6th and 9th terms: simple difference series with constant difference of 3)

Now let’s practice with some IPAT like questions

Q) 7, 6, 14, 12, 21, 18, ?

A] 32

B] 28

C] 14

D] 17

E] 21

Solution

**Solution – B] 28**. Breaking the series into subseries based on 1 term difference we observe-

7, 14, 21, ? (This is 7×1, 7×2, 7×3)

6, 12, 18 (This is 6×1, 6×2, 6×3)

It is seen that terms in each subseries are multiple of the first number in the subseries. Hence the solution is 7×4 = 28

Q) 2, 3, 4, 4, 9, 16, 16, 81, ?

A] 169

B] 144

C] 256

D] 250

E] 181

Solution

**Solution – C] 256**. Breaking the series into three subseries with a 2 term (as 1 term does not yield any solution) interleave difference we see.

2, 4, 16

3, 9, 81

4, 16, ?

We observe the next term in each subseries in the square of the previous term in the same subseries.

Hence the solution is 16×16 = 256

Q) 1, 2, 3, 4, 1.1, 3, 6, 4, 1.21, 4.5, 12, 4, 1.331, ?

A] 6.75

B] 1.441

C] 14.11

D] 7.25

E] 8.25

Solution

**Solution – A] 6.75**. Here is a more complex series, breaking the series into subseries with a 3 term interleave difference we see –

1, 1.1, 1.21, 1.331 ( each next term is formed by multiplying previous term by 1.1)

2, 3, 4.5, ? (each next term is formed by multiplying previous term by 1.5)

3, 6, 12 (each next term is formed by multiplying previous term by 2)

4, 4, 4 (each next term is formed by multiplying previous term by 1)

Consequently, the next term in the series is 4.5×1.5 = 6.75

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