In a geometric progression of positive terms, the 5th term is 9 times the 3rd term and the sum of the 6th and 7th terms is 972. Find the a) common ratiob) sum of the first 6 terms

Answer:a) 3b) 364Step-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.