Given the geometric progression 625,125,25 find the sixth term and the sum of the 15 terms. How would i do this?

Geometrical progression is given by the following formula:

a_n+1=a_n*q, where q is the common ratio. From this formula, you can obtain another one, relating the first term of progression to the arbitrary one,

a_n=a₁*q^n-1;

So, for the sixth term in your case, we need to determine q and a₁. a₁=625; q=125/625, ratio of two successive terms, q=1/5.

a₆=625*(1/5)⁵=625/5⁵=1/5;

Now, for the sum of N terms, from the first to the N^th one.

S_N=a₁+a₂+…+a_N=a₁+a₁*q+a₁*q²+…+a₁*q^N-1=a₁*(1+q+…+q^N-1)

Let us multiply and divide this expression by (q-1). We will obtain the following:

S_N=a₁*(q-1)*(1+q+…+q^N-1)/(q-1)=a₁*(q^N-1)/(q-1)

Thus, for the sum of 15 terms we will get:

S₁₅=625*((1/5)¹⁵-1)/(1/5-1)=625*(1-(1/5)¹⁵)/(4/5)=

=3125/4*(1-(1/5)¹⁵)=3125*30517578124/4/30517578125=7629394531/9765625=781.2499999744

Hello Barbara,

The common ratio of a known geometric sequence can be found by

r = a_n+1/a_n

for some valid n.  In your case, of course, the common ratio turns out to be

r = 125/625 = 1/5.

Now, the formula you presented is not the correct formula for a geometric sequence.  (It is in fact the recursive definition of the Fibonacci  sequence.)  The explicit formula for the n^th term of a geometric sequence is

a_n = a₁r^n-1.

Here, I am assuming that the sequence begins with index 1, i.e. a₁ is the first term.

Since you know your first term (a₁ = 625) and your common ratio (r = 1/5), you are now in position to compute the sixth term:

a₆ = a₁r^6-1 = 625(1/5)⁵.

To find the sum of the first n terms of a geometric sequence, you use the following formula:

S_n = (a₁(1-rⁿ))/(1-r).

You can surely complete the computation on your own.

Regards,

Hassan H.