# Determine the sum of the first 20 terms of an A.P. if a4 - a2 = 4 and a1 + a3 + a5 + a6 = 30.

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The sum is:

Sn = (a1+ an)n/2

We need the sun of the first 20 terms, then we will substitute with n= 20

==> s20 = (a1+ a20) * 20/2

==> S20 = 10(a1+ a20)

Given:

a4-a2= 4

a1+ a3+ a5+a6= 30

We know that: if r is the common ratio between numbers:

a2= a1+ r

a4= a1+ 3r

a5= a1+ 4r

a6= a1+ 5r

but: a4- a2= 4

==> a1+ 3r - (a1+r) = 4

==> 2r = 4

==>** r= 2**

Also:

a1+ a3+ a5+ a6 = 30

==> a1+ a1+2r + a1+ 4r + a1 5r = 30

==> 4a1 + 11r = 30

==> 4a1+ 11*2 = 30

==> 4a1 + 22 = 30

==> 4a1 = 8

==> a1= 8/4= 2

**==> a1= 2**

Now we will calculate a20:

a20 = a1+ 19*r

= 2 + 19*2= 2 + 38 = 40

==> **a20 = 40**

**==> S20 =10 (a1+ a20)**

** = 10( 2 + 40)**

** = 10* 42 = 420 **

**==> S20 = 420 **

We'll write, for the beginning, the sum of n terms of an arithmetic series:

Sn = (a1+an)*n/2

a1 - the 1st term

an - the n-th term

n - the number of terms

Since n = 20, we'll re-write the sum for the first 20 terms:

S20 = (a1 + a20)*20/2

S20 = (a1 + a20)*10

We'll have to calculate the first term and the common difference d, to determine any term of the arithmetic series.

From enunciation, we have:

a4 - a2 = 4

a4 = a1 + 3d

a2 = a1 + d

We'll write a4 and a2 with respect to a1 and d:

a1 + 3d - a1 - d = 4

We'll combine and eliminate like terms:

2d = 4

d = 2

We also know, from enunciation, that:

a1 + a3 + a5 + a6 = 30

We'll write the terms with respect to a1 and d:

a1 + a1 + 2d + a1 + 4d + a1 + 5d = 30

We'll combine like terms and substitute d:

4a1 + 11d = 30

4a1 = 30 - 11d

4a1 = 30 - 22

4a1 = 8

a1 = 2

Now, we can calculate a20:

a20 = a1 + 19d

a20 = 2 + 19*2

a20 = 2 + 38

a20 = 40

S20 = (a1 + a20)*10

S20 = (2 + 40)*10

S20 = 42*10

**S20 = 420**

**The sum of the first 20 terms of the arithmetic progression is S20 = 420.**