Solution:

We agree that the series is quite difficult to crack and much tricky as well.

1 can be written as one (consist of 3 letters)
2 can be written as two (consist of 3 letters)
3 can be written as five (consist of 5 letters)
4 can be written as four (consist of 4 letters)
5 can be written as five (consist of 5 letters)
6 can be written as six (consist of 3 letters)

Thus the next term in the series is 3.

Solution:

This one is a tricky series and here we shall tell you a common way to solve the series problem when fractions are involved. You will find that so many series can be solved the way this one is solved.

Just look at the numerator and denominator separately

Numerators = 1, 3, 5, 7
Two is added to each one, thus the next number will be 9.

Denominators = 1, 2, 4, 8
Every next term is the double of previous, thus the next term will be 16.

So, the next term in the series will be 9/16.

Solution:

This is a difficult one to crack as there are two series in the same series.

One is 2, 10 (2*5), 40 (2*5*4)
The other is 1, 4 (1*4), 16 (4*4), 64 (16*4)

Since we have to find just the next term, it will be linked with the 1, 4, 16, 64 series
And the next term will be (64*4) or 256.

Solution:

Let us assume that the xth term is zero

As per the question,
The 15th term must be (x + 12 d)
6th term be (x + 3d)
11th term be (x + 8d)
13th term be (x + 10d)

Now by the condition given,

2x + 12 d = 3x + 21d
X + 9d = 0

Now x + 9d will be the 12th term

Thus the 12th term in such an arithmetic progression will be 0.

Solution:

This is a bit tricky series problem. Let us break that down for you

23 is the first term
25 = 21*1 + 2
53 = 25*2 + 3
163 = 53*3 + 4
657 = 163*4 + 5
3291 = 657*5 + 6

Thus the next term will be 3291 * 6 + 7 = 19753

Solution:

The first number divisible by 7 is 105 and the last is 994 (we have to find the sum of only three digit numbers)

The divisible numbers are 105, 112, 119, ..., 994
Sum of digits = 105 + 112 + 119, ..., + 994

Now if you see the above series it is an arithmetic progression
Where a = 105, d = 7, I = 994

Thus
994 = 105 + (n - 1)*7
or 889 = (n-1)*7
or 127 = n - 1
or n = 128
so sum = (n/2)*(a + l)
= (128/2)*(105 + 994)
= 70336

Solution:

1595

13
44 = 13 + 31
88 = 44 + 44
176 = 88 + 88
847 = 176 + 671
1595 = 847 + 748

Solution:

Let the four nos. in A.P. are (a-d), (a), (a + d) and (a + 2d).
Now as per the given conditions
4a + 2d = 28
2a+d=14

Also a*(a + 2d) / ((a + d)*(a - d)) = 8/15
15a^2 + 30ad = 8a^2 - 8d^2
7a^2 + 30ad + 8d^2 = 0

Or (a + 4d) (7a + 2d) = 0
a + 4d = 0 or 7a + 2d = 0
Solving a + 4d = 0 and 2a + d = 0 we obtain a = 8 and d = (-2)

Thus the numbers are
4, 6, 8 and 10.

Solution:

The next two numbers will be 42 and 47.
If you look closely, you will see that the following numbers are either a multiple of 7 or the next digit with a 7 in it.

Solution:

The answer is -113

How?
-3*1 - 1 = -4
-4*2 - 1 = -9
-9*3 - 1 = 28
-28*4 - 1 = -113