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Question 0403b
Arithmetic and Geometric Progressions (APs, GPs)
Proofs
Question
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A series is such that the sum of the first
n
{n}
n
terms of the series is given by
S
n
=
1
−
(
1
3
)
n
.
S_n = 1-\left(\frac{1}{3}\right)^n.
S
n
=
1
−
(
3
1
)
n
.
How do we prove that the series is geometric?
Attempt
Select the correct option:
S
n
−
S
n
−
1
=
constant
{S_n - S_{n-1} = \textrm{constant}}
S
n
−
S
n
−
1
=
constant
u
n
−
u
n
−
1
=
constant
{u_n - u_{n-1} = \textrm{constant}}
u
n
−
u
n
−
1
=
constant
S
n
S
n
−
1
=
constant
{\displaystyle \frac{S_n}{S_{n-1}} = \textrm{constant}}
S
n
−
1
S
n
=
constant
u
n
u
n
−
1
=
constant
{\displaystyle \frac{u_n}{u_{n-1}} = \textrm{constant}}
u
n
−
1
u
n
=
constant
By observing the pattern for
S
1
,
S
2
,
S
3
,
…
{S_1, S_2, S_3, \ldots}
S
1
,
S
2
,
S
3
,
…
By observing the pattern for
u
1
,
u
2
,
u
3
,
…
{u_1, u_2, u_3, \ldots}
u
1
,
u
2
,
u
3
,
…
Answer
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