/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 33 Given sum of the first \(n\) ter... [FREE SOLUTION] | 91Ó°ÊÓ

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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) \(3 n+2 n^{2}\)

Short Answer

Expert verified
The sum of n terms of the new A.P. is \(6n^2 - 4n\), corresponding to option (b).

Step by step solution

01

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 \(d\).
02

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 \(d = 6\).
03

Form the New A.P.

In the new A.P., the first term remains \(a = 2,\) but the common difference is doubled: \(2d = 12.\)
04

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:\[ S_n = \frac{n}{2} (4 + 12n - 12) = \frac{n}{2} (12n - 8) = n(6n - 4) \].
05

Simplify Final Expression

Further simplify the expression:\[ n(6n - 4) = 6n^2 - 4n \].
06

Match with Given Options

Compare the simplified expression with the given options:The expression \(6n^2 - 4n\) matches option (b): \(6n^2 - n\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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:
  • \( S_n = \frac{n}{2} (2a + (n-1)d) \)
Here, \(S_n\) is the sum of the first \(n\) terms, \(a\) is the first term, \(d\) is the common difference, and \(n\) is the number of terms.
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:
  • \( d = a_{n+1} - a_n \)
where \(a_{n}\) and \(a_{n+1}\) are consecutive terms of the sequence.
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.
  • Example of general term formula: \( a_n = a + (n-1)d \)
In our exercise, the first term was extracted from the given sum formula by identifying coefficients. We determined \(a = 2\), showing how essential it is to break down the sum formula to solve for \(a\).
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.

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Most popular questions from this chapter

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 April 23, 2013] (a) \(10 \mathrm{~m}\) (b) \(12 \mathrm{~m}\) (c) \(13 \mathrm{~m}\) (d) \(15 \mathrm{~m}\)

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 \(150^{\text {th }}\) term of this AP is : (a) \(-150\) (b) 150 times its \(50^{\text {th }}\) term (c) 150 (d) Zero

Let \(\alpha\) and \(\beta\) be the roots of equation \(\mathrm{px}^{2}+\mathrm{qx}+\mathrm{r}=0, \mathrm{p} \neq 0\). If \(\mathrm{p}, \mathrm{q}, \mathrm{r}\) are in A.P and \(\frac{1}{\alpha}+\frac{1}{\beta}=4\), then the value of \(|\alpha-\beta|\) is: (a) \(\frac{\sqrt{34}}{9}\) (b) \(\frac{2 \sqrt{13}}{9}\) (c) \(\frac{\sqrt{61}}{9}\) (d) \(\frac{2 \sqrt{17}}{9}\)

Let \(a_{1}, a_{2}, a_{3}, \ldots\) be \(a \mathrm{G} .\) P. such that \(a_{1}<0, a_{1}+a_{2}=4\) and \(a_{3}+a_{4}=16 .\) If \(\sum_{i=1}^{9} a_{i}=4 \lambda\), then \(\lambda\) is equal to: [Jan. 7, 2020 (II)] (a) \(-513\) (b) \(-171\) (c) 171 (d) \(\frac{511}{3}\)

For \(x \in \mathrm{R}\), let \([x]\) denote the greatest integer \(\leq x\), then the sum of the series \(\left[-\frac{1}{3}\right]+\left[-\frac{1}{3}-\frac{1}{100}\right]+\left[-\frac{1}{3}-\frac{2}{100}\right]+\cdots+\left[-\frac{1}{3}-\frac{99}{100}\right]\) is [April 12, 2019 (I)] (a) \(-153\) (b) \(-133\) (c) \(-131\) (d) \(-135\)

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