7,9,11,13,15,...?

I'll show you several ways:

Here's the crude way.  Write them out:

 7,  9, 11, 13, 15, 17, 19, 21, 23, 25,
27, 29, 31, 33, 35, 37, 39, 41, 43, 45,
47, 49, 51, 53, 55, 57, 59, 61, 63, 65,
67, 69, 71, 73, 75, 77, 79, 81, 83, 85,
87, 89, 91, 93, 95, 97, 99,101,103,105.

So we know the answer is 105.

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Or we can do it another way:

7,9,11,13,15,...?

Subtract 5 from them all to get them down to
a list of the even numbers:

2,4,6,8,10,,...?

Divide them all by 2

1,2,3,4,5,...?

Now we know the 50th term of that is 50. So we
can put 50 for the question mark there:

1,2,3,4,5,...,50

So we undo what we did to them in reverse order, 
and see what happens to the 50.

The last thing we did was to divide by 2,
so we undo that division by multiplying them all by 2:

2,4,6,8,10,...,100

Before that we subtracted 5 to get them down to the 
even numbers, so we undo that by adding 5 to them all:

7,9,11,13,15,...105 

So again, we see, it's 105.

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Here's another way 

The first term is a1
To get the 2nd term, we add the common difference d 1 time  
to the 1st term a1. a2 = a1 + (1)d
To get the 3rd term, we add the common difference d 2 times 
to the 1st term a1. a3 = a1 + (2)d
To get the 4th term, we add the common difference d 3 times 
to the 1st term a1. a4 = a1 + (3)d
...
To get the nth term, we add the common difference d n-1 times 
to the 1st term a1. an = a1 + (n-1)d
 
an = a1 + (n-1)d

That's where the formula comes from.

In your problem,

7,9,11,13,15,...?

to get the 50th term, add common difference 2, 49 times to the 1st term
a1 = 7. 

an = a1 + (n-1)d
a50 = 7 + (50-1)2
a50 = 7 + (49)2
a50 = 7 + 98
a50 = 105

Edwin