Theory of Geometric Series

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What is Geometric Progression (GP)?

Geometric Series is a series in which consecutive terms have the same ratio.
It is also known as Geometric Sequence or Geometric Series
Example: 2 , 6 , 18 , 54…. ( Consecutive terms have the same ratio of 3:1 )

GP Formulas

\(N^{th}\) Term of a Geometric Series

Geometric Series is given by

\(a , ar, ar^2, ar^3, … , ar^{n-1}\)
\(T_{1}\) = a = \(ar^{1-1}\)
\(T_{2}\) = ar = \(ar^{2-1}\)
\(T_{3}\) = a\(r^2\) = \(ar^{3-1}\)
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\(T_{n}\) = \(ar^{n-1}\)

Sum of n terms of a GP is given by

\(S_{n} = a*\frac{((𝒓^𝒏 − 𝟏)}{(𝐫−𝟏))}\)

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Problem Solution

\(a_1,a_2,8,16,.......,128\) is a geometric sequence

Common Ratio, r = \(\frac{16}{8}\) = 2

Quantity A = \( a_1+a_2\)

=> Let first term be a
=> Third term = a\(r^2\) = 8 (given)
=> a = \(\frac{8}{r^2}\) = \(\frac{8}{2^2}\) = \(\frac{8}{4}\) = 2
=> \(a_1\) = 2
=> \(a_1\) = 2*r = 2*2 = 4
=> Quantity A = \( a_1+a_2\) = 2 + 4 = 6

Quantity B = The number of terms in the sequence : Using two methods

Method 1: Using Logic

Series is 2, 4, 8, ...., 128
=> Series is \(2^1\), \(2^2\), \(2^3\)....., \(2^7\)
=> There are 7 terms in the series

Method 2: Using Formula

Last term of the sequence, \(T_{n}\) = 128
=> \(ar^{n-1}\) = 128
=> \(2*2^{n-1}\) = 128
=> \(2^1*2^{n-1}\) = \(2^7\)
=> \(2^{1+(n-1)}\) = \(2^7\)
=> \(2^n\) = \(2^7\)
=> n = 7

Clearly, Quantity B (7) > Quantity A (6)

So, Answer will be B
Hope it helps!

Watch the following video to learn more about Arithmetic and Geometric Series and Sequences in General