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

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hala718's profile pic

hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted on

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

Top Answer

teacher1955's profile pic

teacher1955 | College Teacher | (Level 1) eNoter

Posted on

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.

 

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

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

neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted on

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.

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