Full Question:
The 4th term of a g.p. is 40 and the 10th term in the sequence is 2560, what is the 11th term in the sequence ?
Answer:
the 11 the term is 5120
Explanation:
Given
Geometry Progression
4th term = 40
10th term = 2560
Required
11 term.
The nth term of a geometric sequence is calculated as follows
Tₙ = arⁿ⁻¹
For the 4th term, n = 4 and Tₙ = 40
Substitute these in the given formula; this gives
40 = ar⁴⁻¹
40 = ar³. --;; equation 1
For the 10th term, n = 10 and Tₙ = 2560
Substitute these in the given formula; this gives
2560 = ar¹⁰⁻¹
2560 = ar⁹. --;; equation 2
Divide equation 2 by 1. This gives
2560/40 = ar⁹/ar³
64 = r⁹/r³
From laws of indices
64 = r⁹⁻³
64 = r⁶
Find 6th root of both sides
(64)^1/6 = r
r = (2⁶)^1/6
r = 2
Substitute r = 2 in equation 1
40 = ar³. Becomes
40 = a * 2³
40 = a * 8
40 = 8a
Divide both sides by 8
40/8 = 8a/8
5 = a
a = 5.
Now, the 11 term can be solved using Tₙ = arⁿ⁻¹ where n = 11
So,
Tₙ = arⁿ⁻¹ becomes
Tₙ = 5 * 2¹¹⁻¹
Tₙ = 5 * 2¹¹⁻¹
Tₙ = 5 * 2¹⁰
Tₙ = 5 * 1024
Tₙ = 5120.
Henxe, the 11 the term is 5120