Problem 12

Question

Let \(\mathrm{S}_{\mathrm{n}}\) denote the sum of the first \(n\) terms of an A.P. If \(\mathrm{S}_{4}=16\) and \(S_{6}=-48\), then \(S_{10}\) is equal to : [April 12, 2019 (I)\\} (a) \(-260\) (b) \(-410\) (c) \(-320\) (d) \(-380\)

Step-by-Step Solution

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Answer

The value of \(S_{10}\) is -320.

1Step 1: Understanding the Formula for Sum of an A.P.

The sum of the first n terms of an arithmetic progression (A.P.) is given by the formula \(S_n = \frac{n}{2} \times (2a + (n-1)d)\), where \(a\) is the first term and \(d\) is the common difference.

2Step 2: Set Up Equations for Given Information

Use the information \(S_4 = 16\) and \(S_6 = -48\) to create two equations using the sum formula:- For \(S_4\): \(\frac{4}{2}(2a + 3d) = 16\), simplifying to \(2a + 3d = 8\).- For \(S_6\): \(\frac{6}{2}(2a + 5d) = -48\), simplifying to \(3(2a + 5d) = -48\), hence \(2a + 5d = -16\).

3Step 3: Solve for Variables a and d

We now have two equations:1. \(2a + 3d = 8\)2. \(2a + 5d = -16\)Subtract the first from the second to eliminate \(2a\): \((2a + 5d) - (2a + 3d) = -16 - 8\)\(2d = -24\), so \(d = -12\).Substitute \(d = -12\) back into \(2a + 3(-12) = 8\):\(2a - 36 = 8\), so \(2a = 44\), thus \(a = 22\).

4Step 4: Find \(S_{10}\) Using the Derived Values

Now, we calculate \(S_{10}\) using \(a = 22\) and \(d = -12\):\[S_{10} = \frac{10}{2} (2a + 9d)\]\(S_{10} = 5(2 \times 22 + 9(-12))\)\(S_{10} = 5(44 - 108)\)\(S_{10} = 5(-64)\)\(S_{10} = -320\).

Key Concepts

Sum of n termsCommon differenceFirst term

Sum of n terms

When exploring arithmetic progressions (A.P.), one of the fundamental concepts is the calculation of the sum of the first \( n \) terms, often denoted as \( S_n \). This sum can be calculated using the formula:\[ S_n = \frac{n}{2} \times (2a + (n-1)d) \]Where:

\( n \) is the total number of terms.
\( a \) is the first term of the sequence.
\( d \) is the common difference between consecutive terms.

This equation allows us to quickly determine the cumulative value of an arithmetic progression up to any specific term. It involves calculating the average of the first and last terms based on the common difference and then extending this average over \( n \) terms. Understanding this formula is important because it shows how the general behavior of the sequence can be determined only by a few known parameters.

Common difference

In arithmetic progressions, the common difference \( d \) is a crucial concept. It represents the constant difference between successive terms of the sequence. Mathematically, it can be expressed as the difference between any two successive terms, such as \( a_2 - a_1 = d \).
Here’s why the common difference is important:

It defines the direction (increasing or decreasing) and the uniform rate of change of the sequence.
It's used in the sum formula for the progression, playing a key role in affecting the sum for any number of terms \( n \).
A positive \( d \) indicates an increasing sequence, while a negative \( d \) denotes a decreasing one.

To visualize the impact of \( d \), imagine how a simple step in a staircase might symbolize the equal spacing brought by \( d \). If the step is higher or lower, you move upward or downward in the overall sequence. This constancy in the difference allows for the prediction of future terms and the sum.

First term

The first term \( a \) of an arithmetic progression is where the journey begins in the sequence. Essentially, it serves as the starting point from which all subsequent terms are calculated.
Why is the first term so essential?

It sets the baseline for the entire sequence. Any deviation in \( a \) would shift the entire progression.
It's integral in calculating the sum of \( n \) terms using the formula \( S_n = \frac{n}{2} \times (2a + (n-1)d) \).
Knowing \( a \), alongside \( d \), helps in formulating the general term expression for the sequence, which is \( a_n = a + (n-1)d \).

If you think of \( a \) as the seed planted, it influences how the whole series of numbers develops. Its value early in the sequence greatly impacts subsequent calculations for things like the sum of terms and the predictability of the sequence's behavior.

Problem 11

Problem 13

Other exercises in this chapter

Problem 10

Let \(f: R \rightarrow R\) be such that for all \(x \in R,\left(2^{1+x}+2^{1-x}\right), f(x)\) and \(\left(3^{x}+3^{-x}\right)\) are in A.P., then the minimum v

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Problem 11

Five numbers are in A.P., whose sum is 25 and product is 2520 . If one of these five numbers is \(-\frac{1}{2}\), then the greatest number amongst them is: [Jan

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Problem 13

If \(a_{1}, a_{2}, a_{3}, \ldots \ldots\) are in A.P. such that \(a_{1}+a_{7}+a_{16}=40\), then the sum of the first 15 terms of this A.P. is : [April 12, 2019

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Problem 14

If \(a_{1}, a_{2}, a_{3}, \ldots . . a_{n}\) are in A.P. and \(a_{1}+a_{4}+a_{7}+\ldots+a_{16}=114\) then \(a_{1}+a_{6}+a_{11}+a_{16}\) is equal to : \(\quad\)

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