Problem 157 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

Show that the sum of \(n\) terms of a G.P. of common ratio \(r\) beginning with the \(p^{t h}\) term is \(r^{p-q}\) times the sum of an equal number of terms of the same series beginning with \(q^{t h}\) term.

Step-by-Step Solution

Answer

The sum of \(n\) terms of a Geometric Progression of common ratio \(r\) beginning with the \(p^{th}\) term is indeed \(r^{p-q}\) times the sum of the same number of terms starting from the \(q^{th}\) term.

Step 1: Define a term of the series

First, define an arbitrary term of the geometric sequence. If the first term of the general geometric progression is 'a' and the common ratio is 'r', then any term \(P_n\) can be expressed as: \(P_n = ar^{(n-1)}\) and \(Q_m = ar^{(m-1)}\), where \(P_n\) and \(Q_m\) are the \(n^{th}\) and \(m^{th}\) term of the GP respectively.

Step 2: Find sums of geometric progressions

Next, calculate the sum of the first n terms of a geometric progression starting with the \(p^{th}\) term and the sum starting with the \(q^{th}\) term. So, we have the sums \(S_{n, p} = P_p + P_{p+1} + P_{p+2} + ... + P_{p + (n - 1)}\) and \(S_{n, q} = Q_q + Q_{q+1} + Q_{q+2} + ... + Q_{q + (n - 1)}\) for the pth and qth starting terms respectively.

Step 3: Apply formulas for the sum of geometric series

Now, recall that the sum of \(n\) terms of a geometric series with first term \(a\) and ratio \(r\) is given by the formula \(S = \frac{a(r^n - 1)}{r - 1}\). Applying this formula, the sums become \(S_{n, p} = \frac{P_p(r^n - 1)}{r - 1}\) and \(S_{n, q} = \frac{Q_q(r^n - 1)}{r - 1}\)

Step 4: Link the sums of the series

Notice that we can express \(P_p\) as \(P_p = Q_q * r^{p - q}\), which comes from the initial definition of any term in our series. This means we can replace \(P_p\) in \(S_{n, p}\) formula: \(S_{n, p} = \frac{P_p(r^n - 1)}{r - 1} = \frac{(Q_q * r^{p - q})(r^n - 1)}{r - 1}\). Simplifying this we get \(S_{n, p} = r^{p-q} * S_{n, q}\), which is the result we wanted to prove.

Step 5: Final Statement

We have proved that the sum of \(n\) terms of a GP of common ratio \(r\) beginning with the \(p^{th}\) term is \(r^{p-q}\) times the sum of an equal number of terms of the same series beginning with the \(q^{th}\) term.

Key Concepts

Sum of n Terms

In a geometric progression (G.P.), the sum of the first \( n \) terms is a crucial calculation. If you are dealing with such a series, it begins with a first term 'a' and scales by the common ratio 'r' each time. The formula to find this sum is given by: \[ S = \frac{a(r^n - 1)}{r - 1} \] This formula allows us to easily compute the sum by substituting 'a' with the first term of the sequence, 'r' with the common ratio, and 'n' with the number of terms you want to sum. In our exercise, this sum is calculated starting from specific terms labeled as the \( p^{th} \) and \( q^{th} \) terms, showing the flexibility of the formula to not just start from the beginning.

Common Ratio

The common ratio in a geometric progression is the factor by which each term is multiplied to get to the next term. It is a constant value throughout the sequence. For instance, in a G.P. like 2, 6, 18, 54, and so on, the common ratio is 3 as each number is 3 times the previous number. This ratio 'r' is a critical part of both the sequence itself and in calculating the sum of the series. In the given exercise, understanding how the ratio 'r' operates helps demonstrate why multiplying one sum by \( r^{p-q} \) aligns the sums of sequences beginning at different terms.

Series Comparison

Comparing two series in a geometric progression may involve inspecting sums starting at different points. Specifically, we examine their sums starting from the \( p^{th} \) and \( q^{th} \) terms.In the exercise, the comparison showed the sum of n terms beginning with the \( p^{th} \) term is \( r^{p-q} \) times the sum starting from the \( q^{th} \) term. This comparison underscores not only the power of geometric progression formulas but also how the common ratio 'r' scales the sums of different parts of the series effectively.

Geometric Series

A geometric series comprises the sum of the terms of a geometric progression. It is one of the simplest forms of a series, characterized by its first term and a constant ratio.Learning about geometric series is vital because it provides insight into how sequences behave under multiplication by a constant factor. In the context of the exercise, acknowledging that the series transformation through multiplication by \( r^{p-q} \) will allow us to present a succinct relationship between sums starting at different terms, illuminating the predictable nature of geometric patterns.