# Find the sum of the first 4 terms of a geometric progression if a3 = 1 and the common ratio between consecutive terms is 1/2.

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### 2 Answers

The consecutive terms of a GP have a common ratio which is given here as (1/2). Let the first term of the GP be a.

The nth term is given as Tn = a*r^(n - 1)

T3 = 1 = a*(1/2)^2

=> a = 4

The sum of the first terms of a GP where r is less than 1 is given by Sn = a*(1 - r^n)/(1 - r)

S4 = 4*(1 - (1/2)^4)/(1 - (1/2))

=> 4*(1 - 1/16)/(1/2)

=> 4*2[(16 - 1)/16]

=> 8*15/16

=> 15/2

**The sum of first 4 terms is 15/2**

Let the terms of a geometric progression be :

a1, a2, a3, a4

Given that a3= 1 and the common difference is 1/2.

==> Then we know that:

a3 = a1*r^2

1 = a1* (1/2)^2

1= a1/ 4

==> a1= 4

==> a2= a1*r = 4* 1/2 = 2

==> a3= a1*r^2 = 4*1/4 = 1

==> a4= a1*r^3 = 4*(1/8) = 1/2

Then the terms are:

4, 2, 1, 1/2

We will find the sum.

==> S = 4 + 2 + 1 + 1/2 = 7.5

**Then the sum of the first 4 terms is 7.5**