Find a1 for the geometric sequence with the given terms. a5 = 81 and 17 = 729

Answer 1

a₁r⁴ = 81; a₁r¹⁶ = 729; a₁⁴r¹⁶ = 3¹⁶; a₁r¹⁶ = 3⁶; a₁³ = 3¹⁰; a₁ = 3^10/3

Answer 2

think you mean a₇=729

then r=+/-3

sequence is {kr⁰,kr¹,...,kr⁷}

kr⁰=a₁

know a₅=81

and a₅=k*(+/-3)⁴=k*81

k=1

a₁=1*3⁰=1

Answer 3

a5 = 81

a17 = 9(81) =729

an = a1(r^(n-1))

a5 =81 = a1(r(4))

a17 = 729 = a1(r^16))

divide

729/81 =r^(16-4) = r^12

9 = r^12

r = twelfth root of 9 = about 1.2

a1 = 81/r^4 = 81/(9^(4/12) = 81/(9^1/3)=about 81/2.08

a1 = about 38.9407384

a1 = about 39

IF a7= 729 = 9(81)

a5 = 81

then a3 = 9

a1 = 1

1,3,9,27,81,243, 729, ...

common ratio =r = 3

an=a1(r^(n-1))

a5=a1(3^4)

81 =a1(81)

a1 =81/81=1