Answer:
a) 3
b) 364
Step-by-step explanation:
A geometric sequence in explicit form is [tex]a_n=a_1 \cdot r^{n-1}[/tex] where [tex]a_1[/tex] is the first term and [tex]r[/tex] is the common ratio.
We are given:
[tex]a_5=9 \cdot a_3[/tex]
[tex]a_6+a_7=972[/tex].
What is a) r?
What is b) the sum of the first 6 terms?
So I'm going to use my first equation and use my explicit form to find those terms in terms of r:
[tex]a_1 \cdot r^4=9 \cdot a_1 \cdot r^{2}[/tex]
Divide both sides by [tex]a_1r^2[/tex]:
[tex]r^2=9[/tex]
[tex]r=\sqrt{9}[/tex]
[tex]r=3[/tex].
So part a is 3.
Now for part b).
We want to find [tex]a_1+a_2+a_3+a_4+a_5+a_6[/tex].
So far we have:
[tex]a_1=a_1[/tex]
[tex]a_2=3a_1[/tex]
[tex]a_3=3^2a_1[/tex]
[tex]a_4=3^3a_1[/tex]
[tex]a_5=3^4a_1[/tex]
[tex]a_6=3^5a_1[/tex].
We also haven't used:
[tex]a_6+a_7=972[/tex].
I'm going to find these terms in terms of r (r=3).
[tex]3^5a_1+3^6a_1=972[/tex]
[tex]243a_1+729a_1=972[/tex]
You have like terms to add:
[tex]972a_1=972[/tex]
Divide both sides by 972:
[tex]a_1=1[/tex]
The first term is 1 and the common ratio is 3.
The terms we wrote can be simplify using a substitution for the first term as 1:
[tex]a_1=a_1=1[/tex]
[tex]a_2=3a_1=3(1)=3[/tex]
[tex]a_3=3^2a_1=9(1)=9[/tex]
[tex]a_4=3^3a_1=27(1)=27[/tex]
[tex]a_5=3^4a_1=81(1)=81[/tex]
[tex]a_6=3^5a_1=243(1)=243[/tex].
Now we just need to find the sum of those six terms:
1+3+9+27+81+243=364.
