Problem 85 - A Complete Resource Book in Mathematics for JEE Main

Question

If, in a G.P. of \(3 n\) terms, \(S_{1}\) denotes the sum of the first \(n\) terms, \(S_{2}\) the sum of the second block of \(n\) terms and \(S_{3}\) the sum of the last \(n\) terms, then \(S_{1}, S_{2}, S_{3}\) are in (A) A.P. (B) G.P. (C) H.P. (D) None of these

Step-by-Step Solution

Answer

B) G.P.

Step 1: Understanding the Problem

We are given a geometric progression (G.P.) with a total of \(3n\) terms. The sequence is divided into three blocks, each containing \(n\) terms, and we denote the sums of these blocks as \(S_1\), \(S_2\), and \(S_3\) respectively.

Step 2: Formula for Sum of Terms in G.P.

The sum of the first \(n\) terms of a G.P. is given by the formula \(S = a \frac{r^n - 1}{r - 1}\) where \(a\) is the first term and \(r\) is the common ratio.

Step 3: Calculating \(S_1\)

\(S_1\) is the sum of the first \(n\) terms of the G.P., so \(S_1 = a \frac{r^n - 1}{r - 1}\).

Step 4: Calculating \(S_2\)

The second group of \(n\) terms in the G.P. starts from the \((n+1)\)th term. Since each term in the G.P. is the first term multiplied by a power of \(r\), the first term of the second block is \(ar^n\). Therefore, \(S_2 = ar^n \frac{r^n - 1}{r - 1}\).

Step 5: Calculating \(S_3\)

Similarly, the third block starts from the \((2n+1)\)th term, making the first term of this block \(ar^{2n}\). Thus, \(S_3 = ar^{2n} \frac{r^n - 1}{r - 1}\).

Step 6: Identifying the Relationship

Now that we have \(S_1, S_2, S_3\) as follows:\(S_1 = a \frac{r^n - 1}{r - 1}\), \(S_2 = ar^n \frac{r^n - 1}{r - 1}\), \(S_3 = ar^{2n} \frac{r^n - 1}{r - 1}\).Notice that \(S_2 = S_1 \cdot r^n\) and \(S_3 = S_2 \cdot r^n\), demonstrating that \(S_1, S_2, S_3\) forms a G.P.

Key Concepts

Sum of Geometric Series

Understanding the sum of a geometric series is crucial in mathematics, especially when dealing with sequences where each term is a constant multiple of the previous one. In a geometric progression (G.P.), the sum of the first \( n \) terms can be calculated using the formula:

\[S_n = a \frac{r^n - 1}{r - 1}\]where:

• \( S_n \) is the sum of the first \( n \) terms
• \( a \) is the first term
• \( r \) is the common ratio

This formula is derived from the property that each term is created by multiplying the previous term by \( r \). To find the sum of a specific number of terms, the formula helps you factor in both the starting point \( a \) and the growth factor \( r \).

In solving problems like the one in the exercise, understanding this formula helps break down the series into manageable chunks. This allows the calculation of sums for different blocks of terms within a larger sequence, as shown when \( S_1, S_2, \) and \( S_3 \) were calculated separately and demonstrated to form another geometric progression.

Arithmetic Progression

An arithmetic progression (A.P.) is a sequence where each term after the first is derived by adding a constant difference to the previous term. Unlike geometric progressions, the amount added, known as the common difference, stays constant.

The general formula for any term in an A.P. is given by:
\[a_n = a + (n-1) imes d\]where:

• \( a_n \) is the \( n \)-th term
• \( a \) is the first term
• \( d \) is the common difference

The simplicity of this progression means each term linearly increases or decreases by the same measure, making predictions and calculations straightforward.

In contrast to the problem at hand, where sums \( S_1, S_2, \) and \( S_3 \) are investigated, an arithmetic sequence's sums wouldn't multiply in the fashion of a geometric progression. Instead, the differences or sums in an A.P. hinge on a steady increment without the exponential factor seen in G.P.s.

Harmonic Progression

Harmonic progression (H.P.) is slightly less straightforward than the other two progressions. A sequence is considered harmonic when it consists of the reciprocals of an arithmetic sequence. If you have an A.P., \( a, a+d, a+2d, \ldots \), converting these into their reciprocals gives you an H.P.: \( \frac{1}{a}, \frac{1}{a+d}, \frac{1}{a+2d}, \ldots \).

Understanding H.P. helps in situations where inverse relationships are involved or when quantities inversely depend on others. Although it's not directly related to the problem solved in the original exercise, recognizing the definition and nature of H.P.s strengthens broader mathematical understanding.

It is seldom applied in routine calculations like evaluating the sum of blocks in a sequence, as it requires a different approach to sums presented in the likes of geometric or arithmetic progressions. However, knowing how harmonic sequences connect to their arithmetic counterparts offers a richer grasp of mathematical sequences overall.