Question

The fourth term of a geometric expreion erie i 27 and the eventh term i 729. Find the firt term and common ratio and the um of the firt ten term

Answer 1

a₁ = 1 , r = 3 , S₁₀ = 29524

Explanation:

the nth term of a geometric sequence is

`a_(n)` = a₁
`r^(n-1)`

where a₁ is the first term and r the common ratio

given a₄ = 27 and a₇ = 729 , then

a₁r³ = 27 → (1)

a₁
`r^(6)` = 729 → (2)

divide (2) by (1)

`(a_(1)r^(6) )/(a_(1)r^(3) )` =
`(729)/(27)` ( cancel a₁ on numerator/ denominator )

r³ = 27 ( take cube root of both sides )

r =
`\sqrt[3]{27}` = 3

substitute r = 3 into (1) and solve for a₁

a₁ × 3³ = 27

27a₁ = 27 ( divide both sides by 27 )

a₁ = 1

the sum to n terms of a geometric sequence is

`S_(n)` =
`(a_(1)(r^(n)-1) )/(r-1)`

then

S₁₀ =
`(1(3^(10)-1) )/(3-1)`

=
`(x59049-1)/(2)`

=
`(59048)/(2)`

= 29524