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

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2 Answers

Geometrical progression is given by the following formula:

an+1=an*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,

an=a1*qn-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 Nth one.

SN=a1+a2+…+aN=a1+a1*q+a1*q2+…+a1*qN-1=a1*(1+q+…+qN-1)

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

SN=a1*(q-1)*(1+q+…+qN-1)/(q-1)=a1*(qN-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 = an+1/an

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 nth term of a geometric sequence is

an = a1rn-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 = a1r6-1 = 625(1/5)5.

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

Sn = (a1(1-rn))/(1-r).

You can surely complete the computation on your own.

Regards,

Hassan H.