nth term of a geometric:
an = a1 rⁿ ⁻ ¹
a1 = initial value
r = common ratio
In this case:
a1 = 54
a4 = 2
a4 = a1 r⁴ ⁻ ¹ = 2
54 ∙ r³ = 2
r³ = 2 / 54
r³ = 2 / 2 ∙ 27
r³ = 1 / 27
r = ∛ (1 / 27 )
r = ∛1 / ∛27
r = 1 / 3
Sn = Sum first n terms of GP
Sn = a1 ( 1 - rⁿ ) / ( 1 - r )
S5 = Sum first 5 terms of GP
S5 = a1 ( 1 - r⁵ ) / ( 1 - r )
S5 = 54 [ 1 - ( 1 / 3 )⁵ ] / ( 1 - 1 / 3 )
S5 = 54 [ 1 - ( 1⁵ / 3⁵ ) ] / ( 3 / 3 - 1 / 3 )
S5 = 54 ( 1 - 1 / 243 ) / ( 2 / 3 )
S5 = 54 ( 243 / 243 - 1 / 243 ) / ( 2 / 3 )
S5 = 54 ( 242 / 243 ) / ( 2 / 3 )
S5 = ( 54 ∙ 242 / 243 ) / ( 2 / 3 )
S5 = 54 ∙ 242 ∙ 3 / 243 ∙ 2
S5 = 39204 / 486
S5 = 162 ∙ 242 / 162 ∙ 3
S5 = 242 / 3
Proof:
a1 = 54
a2 = 54 / 3 = 18
a3 = 18 / 3 = 6
a4 = 6 / 3 = 2
a5 = 2 / 3
a1 + a2 + a3 + a4 + a5 = 54 + 18 + 6 + 2 + 2 / 3 =
80 + 2 / 3 = 240 / 3 + 2 / 3 = 242 / 3
Find the sum to 5 terms of the geometric progression whose first term is 54 and fourth term is 2
1 answer