# I am given a series 3,6,12,24.... What is the sum of the 5th, 6th and 7th terms of this series.

### 4 Answers | Add Yours

3, 6, 12, 25

a1 = 3

a2= 3*2 = 6

a3 = 3*2^2 = 12

a4 = 3*2^3 = 24

We notice that we have a Geometric progression with constant difference r = 2

==> an = 3*r^(n-1)

Now let us calculate 5th, 6th, and 7th terms:

a5 = 3*r^(5-1) = 3*2^4 = 48

a6 = 3*r^(6-1) = 3*2^5 = 96

a7 = 3*r^(7-1) = 3*2^6 = 192

Then sum of ba5, a6, and a7 is:

S = a5 +a6 + a7

= 48 +96 + 192

= 336

First look at the series you have been given: 3, 6, 12, 24…

You will see that any term divided by the previous term gives a constant number: 6/3=12/6=24/12=2. This is a characteristic of a geometric progression. Therefore the series given is a geometric progression. Now the nth term of a geometric progression is given by ar^(n-1), where a is the first term and r is the common ratio. For the series a=3 and r=2.

The easiest way to find the answer to this question is to find the terms 5, 6 and 7, and then add them up. Using the expression of the nth term of the series as 3*2^(n-1), we get the 5th term as 3*2^4=48, the 6th term as 3*2^5=96 and the 7th term as 3*2^6=192.

Adding 48, 96 and 192 we get 336. This is the sum of the 5th, 6th and 7th terms of the series.

For the beginning, we'll note the terms of teh sequence are the terms of a G.P. as:

a, a*r, a*r^2, ..., where a = 3

We can calculate the common ratio, r, in this way:

a = 3

a*r = 6

3*r = 6

r = 2

a*r^2 = 12

3*r^2 = 12

r^2 = 4

r = 2

To calculate the sum of the 5th, 6th and 7th terms, we'll write:

a5 = a*r^4

a6 = a*r^5

a7 = a*r^6

a5 + a6 + a7 = a*r^4 + a*r^5 + a*r^6

We'll factorize and we'll get:

a*r^4(1 + r + r^2 )

3*2^4(1+2+4) = 3*16*7 = 21*16

**a5 + a6 + a7 = 336**

Terms given are : a1 = 3,

a2 = 6.

a3 =12,

a4 =24.

Clearly a2/a1 = a3/a2 = a4/a3 = 2.

Therefore this is a geometric series (GS) with common ratio r =2

The n th term an of the GS = a1*r^(n-1).

So a5 = a1*r^(5-1) = 3*2^(5-1) = 48

a6 = a1*r^(6-1) =3*2^(6-1) = 96

a7 = a1*r^(7-1) = a6 *r = 96*2 = 192.

If the sum Sn up to the nth term is given by:

Sn = a(r^(n+1)-1)/(r-1). = 3{2^(n) - 1}/(2-1) = 2(2^n)-1).

So S5 = 3[2^5 -1] = 93

S6 = 3(2^6-1) = 189

S7 = 3(2^7-1) = 381

This could be cross checked from the series whose actual terms are as below:

So in the Given GS:

3,6,12,24, **48,96,192** , the bold types are 5th 6th and 7th terms.