# What is the result of the sum of the first term and the common ratio of a GP that has the term b6=25 and b8=9 ?

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

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Given b1, b2, ..., b8 are terms of a G.P.

Let the common difference be r.

Then we need to find the sum b1 + r.

Let us simplify.

We know that
b6= 25  But b6 = b1* r^5

==> b1*r^5 = 25 ...............(1)

Also given b8 = 9

==> b8 = b1*r^7 = 9

==> b1*r^7 = 9 ................(2)

Now we will divide (1) by (2).

==> b1*r^5/b1*r^7 = 25/9

==> 1/r^2 = 25/9

==> r^2 = 9/25

==> r= 3/5

Now we will substitute into (1).

==> b1*r^5 = 25

==> b1*(3/5)^5 = 25

==> b1= 25 * 5^5/ 3^5 = 5^7/ 3^5

==> b1= (5^7)/ (3^5)

==> b1+ r = (5^7/3^5) + (3/5)

= (5^8 + 3^6) / 5*3^5

==> b1+ r = (5^8 + 3^6) / 5*3^5 = 322.1 ( approx)

justaguide | College Teacher | (Level 2) Distinguished Educator

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For a GP the 6th term T6 = 25 and the 8th term T8 = 9.

Now if the first term is a and the common ratio is.

Tn = a*r^(n-1)

So, we have

9 = a*r^7

25 = a*r^5

divide the two

=> 9/25 = r^2

=> r = -3/5 or +3/5

substitute them in 9 = a*r^7

=> 9 = a*(3/5)^7

=> a = 9/(3/5)^7

or a = 9/(-3/5)^7

Therefore a + r = -3/5 +9/(-3/5)^7

or a+ r = 3/5 + 9/(3/5)^7

The required sum can be -3/5 +9/(-3/5)^7 and 3/5 + 9/(3/5)^7.

giorgiana1976 | College Teacher | (Level 3) Valedictorian

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We'll write the formula for the general term of any geometric progression:

bn=b1*r^(n-1), where b1 is the first term, r is the common ratio

Since we know the values of the 6th and the 8th terms, from enunciation, we'll substitute them into the formula of the general terms:

b8=b7*r

But b7=b6*r

So, b8=b6*r*r=b6*r^2

9=25*r^2

We'll divide by 25 both sides:

r = sqrt (9/25)

r = 3/5

But the 6th term could be written as:

b6=b1*r^5

25=b1*(3/5)^5

b1=5^7/3^5

b1+r=(5^7/3^5) + (3/5)

b1+r=(5^8 + 3^6)/5*3^5