Geometric progression - sum of 3,6,12,......1536

[nae99]

can some1 plz help me with this, if its even with the formula
3,6,12,......1536
a) determine the number of terms in the progression
B) Hence, obtain the sum of the progression

thanks.

[Guest]

Use the formula for sum of a geometric progression
[tex]Sum = \frac{a_1 - a_nq}{1 - q}[/tex]

[tex]a_1 = 3[/tex]
[tex]a_n = 1536[/tex]
[tex]q = 3[/tex]

[razabashir]

can some1 plz help me with this, if its even with the formula
3,6,12,......1536
a) determine the number of terms in the progression
B) Hence, obtain the sum of the progression





_________________________
raza

[Guest]

You are dealing with Gepmetric Progression
You need to consider the first term......the starting point
You need to consider the ratio between adjacent numbers in the progression....a common ratio.."r"
Is the ratio the same for all adjacent terms.....then it is a geometric progression
You need to consider the Nth term.....1st, 2nd, 3rd,etc.....up to Nth term

Formula for value of the Nth term..... Nth term value = ar^(n-1)

In your question you know the Nth term = 1536 and the 1st term "a" is 3.
You can see the common ratio is 2.....the progression is diverging...getting bigger and bigger..
You question asks you to find the number of terms in the progression up to the Nth term
So simply put this in the formulae......

1536 = 3 x 2^(n-1)
512 = 2^(n-1)
Logtobase2of (512) = (n-1) x Logtobase2of (2)

Logtobase2of (512) = (n-1)
9 = n-1
9 + 1 = n

n = 10

So 1536 is the 10th term in the progression.

Its so simple we can list them...

3, 6, 12, 24, 48, 96, 192, 384, 768, 1536

Now use the formulae for the sum of n terms to find the "total sum" value of all the terms listed.

.........................Simple.................