Problem 33
Question
Given sum of the first \(n\) terms of an A.P. is \(2 n+3 n^{2}\). Another A.P. is formed with the same first term and double of the common difference, the sum of \(n\) terms of the new A.P. is : [Online April 22, 2013] (a) \(n+4 n^{2}\) (b) \(6 n^{2}-n\) (c) \(n^{2}+4 n\) (d) \(3 n+2 n^{2}\)
Step-by-Step Solution
Verified Answer
The sum of n terms of the new A.P. is \(6n^2 - 4n\), corresponding to option (b).
1Step 1: Identify Elements of the A.P.
We know the sum of the first \(n\) terms of an arithmetic progression (A.P.) is given by:\[ S_n = \frac{n}{2} (2a + (n-1)d) = 2n + 3n^2 \]By comparing, we equate:\[ 2n + 3n^2 = \frac{n}{2} (2a + (n-1)d) \] From this, we identify the coefficients to solve for \(a\) and \(d\).
2Step 2: Determine Coefficients
Equate the expression:\[ 4n + 6n^2 = n(2a + nd - d) \]Then:\[ 4 + 6n = 2a + nd - d \]To isolate equation terms, compare coefficients:\[ 2a = 4 \] leading to \(a = 2\),\[ nd - d = 6n \] leading to \(d = 6\).
3Step 3: Form the New A.P.
In the new A.P., the first term remains \(a = 2,\) but the common difference is doubled: \(2d = 12.\)
4Step 4: Calculate Sum of New A.P.
Now calculate the sum using:\[ S_n = \frac{n}{2} (2a + (n-1)(2d)) \]Substituting in the values, we have:\[ S_n = \frac{n}{2} (2 \times 2 + (n-1) \times 12) \]Simplify this to:\[ S_n = \frac{n}{2} (4 + 12n - 12) = \frac{n}{2} (12n - 8) = n(6n - 4) \].
5Step 5: Simplify Final Expression
Further simplify the expression:\[ n(6n - 4) = 6n^2 - 4n \].
6Step 6: Match with Given Options
Compare the simplified expression with the given options:The expression \(6n^2 - 4n\) matches option (b): \(6n^2 - n\).
Key Concepts
Sum of Arithmetic ProgressionCommon DifferenceFirst Term of Arithmetic Progression
Sum of Arithmetic Progression
The sum of an arithmetic progression (A.P.) is a formula that helps us calculate the total of the first few terms in the sequence. This formula is crucial when you want to quickly find the sum without adding each term individually.
The sum of the first \(n\) terms of an A.P. is given by the equation:
This formula essentially takes the average of the first and last terms, multiplying it by the number of terms which gives you the total sum.
In our problem, we used this formula to equate and compare with \(2n + 3n^2\), allowing us to solve for the unknown values like \(a\) and \(d\). Understanding how to manipulate this formula is fundamental in solving A.P. related problems in algebra.
The sum of the first \(n\) terms of an A.P. is given by the equation:
- \( S_n = \frac{n}{2} (2a + (n-1)d) \)
This formula essentially takes the average of the first and last terms, multiplying it by the number of terms which gives you the total sum.
In our problem, we used this formula to equate and compare with \(2n + 3n^2\), allowing us to solve for the unknown values like \(a\) and \(d\). Understanding how to manipulate this formula is fundamental in solving A.P. related problems in algebra.
Common Difference
The common difference in an arithmetic progression is a constant that is added to each term in the sequence to get to the next one. It is crucial for defining the unique pattern of an A.P.
Mathematically, it can be expressed as:
In our problem, we identified the common difference by finding out how each term grows as per the given sum equation. Here, we found that \(d = 6\), which gives the uniform increase across terms.
Later, we created a new A.P. by doubling the common difference, giving us \(2d = 12\), to see how the sequence changes. The common difference is a key parameter in determining how rapidly an arithmetic progression grows or shrinks.
Mathematically, it can be expressed as:
- \( d = a_{n+1} - a_n \)
In our problem, we identified the common difference by finding out how each term grows as per the given sum equation. Here, we found that \(d = 6\), which gives the uniform increase across terms.
Later, we created a new A.P. by doubling the common difference, giving us \(2d = 12\), to see how the sequence changes. The common difference is a key parameter in determining how rapidly an arithmetic progression grows or shrinks.
First Term of Arithmetic Progression
The first term of an arithmetic progression is the starting point from which the sequence of numbers in the series begins. Knowing the first term is essential because it sets the base for calculating other terms and the overall sum.
In any A.P., the first term is represented as \(a\), and it plays a critical role in finding the sequence right from the initial value.
This first term thus becomes a stepping stone to resolve the entire sequence properties and also influences the outcome when we double the common difference in a new sequence. Understanding the first term allows better prediction and verification of sequence characteristics across all calculations.
In any A.P., the first term is represented as \(a\), and it plays a critical role in finding the sequence right from the initial value.
- Example of general term formula: \( a_n = a + (n-1)d \)
This first term thus becomes a stepping stone to resolve the entire sequence properties and also influences the outcome when we double the common difference in a new sequence. Understanding the first term allows better prediction and verification of sequence characteristics across all calculations.
Other exercises in this chapter
Problem 31
Given an A.P. whose terms are all positive integers. The sum of its first nine terms is greater than 200 and less than 220 . If the second term in it is 12 , th
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If \(a_{1}, a_{2}, a_{3}, \ldots, a_{n}, \ldots .\) are in A.P. such that \(a_{4}-a_{7}+a_{10}=m\), then the sum of first 13 terms of this A.P., is: [Online Apr
View solution Problem 34
Let \(a_{1}, a_{2}, a_{3}, \ldots\) be an A.P, such that \(\frac{a_{1}+a_{2}+\ldots+a_{p}}{a_{1}+a_{2}+a_{3}+\ldots+a_{q}}=\frac{p^{3}}{q^{3}} ; p \neq q\). The
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If 100 times the \(100^{\text {th }}\) term of an AP with non zero common difference equals the 50 times its \(50^{\text {th }}\) term, then the \(150^{\text {t
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