The first and the second terms of an AP are $\frac{5}{2}$ and $\frac{23}{12}$ respectively. If n^th term is the largest negative term, what is the value of n ?

Question

The first and the second terms of an AP are $\frac{5}{2}$ and $\frac{23}{12}$ respectively. If nth term is the largest negative term, what is the value of n ?

Prepp Team · Accepted Answer

Finding the Largest Negative Term in an Arithmetic Progression (AP)

An Arithmetic Progression (AP) is a sequence of numbers such that the difference between the consecutive terms is constant. This constant difference is called the common difference, denoted by $d$.

The first term of the AP is given as $a = \frac{5}{2}$.

The second term of the AP is given as $a_2 = \frac{23}{12}$.

The common difference $d$ is calculated as the difference between the second term and the first term:

$d = a_2 - a$

$d = \frac{23}{12} - \frac{5}{2}$

To subtract these fractions, we need a common denominator, which is 12.

$d = \frac{23}{12} - \frac{5 \times 6}{2 \times 6}$

$d = \frac{23}{12} - \frac{30}{12}$

$d = \frac{23 - 30}{12}$

$d = -\frac{7}{12}$

The common difference is $-\frac{7}{12}$. Since the common difference is negative, the terms of the AP are decreasing.

Calculating the nth Term of the AP

The formula for the nth term ($a_n$) of an AP is given by:

$a_n = a + (n-1)d$

Substitute the values of the first term ($a = \frac{5}{2}$) and the common difference ($d = -\frac{7}{12}$):

$a_n = \frac{5}{2} + (n-1)\left(-\frac{7}{12}\right)$

Finding the Largest Negative Term

We are looking for the largest negative term. In a decreasing AP (where $d < 0$), the terms go from positive towards negative. The largest negative term will be the first term that becomes negative.

To find when the terms become negative, we set the nth term to be less than zero:

$a_n < 0$

$\frac{5}{2} + (n-1)\left(-\frac{7}{12}\right) < 0$

Now, we solve this inequality for $n$.

Multiply both sides by 12 to eliminate denominators:

$12 \times \left(\frac{5}{2}\right) + 12 \times (n-1)\left(-\frac{7}{12}\right) < 12 \times 0$

$6 \times 5 + (n-1)(-7) < 0$

$30 - 7(n-1) < 0$

$30 - 7n + 7 < 0$

$37 - 7n < 0$

Add $7n$ to both sides:

$37 < 7n$

Divide by 7:

$\frac{37}{7} < n$

Calculate the value of $\frac{37}{7}$:

$\frac{37}{7} \approx 5.2857$

So, the inequality is $n > 5.2857$.

Since $n$ must be an integer (term number), the smallest integer value of $n$ that is greater than 5.2857 is $n = 6$.

This means the 6th term is the first term that is negative. Since the AP is decreasing, the first negative term is the largest negative term.

Let's verify the 5th and 6th terms:

For $n=5$: $a_5 = \frac{5}{2} + (5-1)\left(-\frac{7}{12}\right) = \frac{5}{2} + 4\left(-\frac{7}{12}\right) = \frac{5}{2} - \frac{28}{12} = \frac{5}{2} - \frac{7}{3} = \frac{15}{6} - \frac{14}{6} = \frac{1}{6}$. This is positive.
For $n=6$: $a_6 = \frac{5}{2} + (6-1)\left(-\frac{7}{12}\right) = \frac{5}{2} + 5\left(-\frac{7}{12}\right) = \frac{5}{2} - \frac{35}{12} = \frac{30}{12} - \frac{35}{12} = -\frac{5}{12}$. This is negative.

$\frac{1}{6} > -\frac{5}{12}$. The term just before the first negative term is positive. The first negative term is $-\frac{5}{12}$.

In a decreasing AP, the first negative term is the largest negative term (closest to 0). Thus, the largest negative term is the 6th term, and $n=6$.

Summary of Calculations

Step	Description	Calculation
1	Identify first term (a)	$a = \frac{5}{2}$
2	Identify second term (a<sub>2</sub>)	$a_2 = \frac{23}{12}$
3	Calculate common difference (d)	$d = a_2 - a = \frac{23}{12} - \frac{5}{2} = -\frac{7}{12}$
4	Set up inequality for negative term	$a_n < 0 \implies a + (n-1)d < 0$
5	Substitute values and solve for n	$\frac{5}{2} + (n-1)\left(-\frac{7}{12}\right) < 0 \implies n > 5.2857$
6	Find smallest integer n satisfying inequality	$n = 6$

The value of $n$ for which the nth term is the largest negative term is 6.

Revision Table: Arithmetic Progression Concepts

Concept	Formula/Description	Notes
First Term	$a$	The initial term of the sequence.
Common Difference	$d = a_k - a_{k-1}$	Constant difference between consecutive terms.
nth Term	$a_n = a + (n-1)d$	Formula to find any term in the sequence.
Largest Negative Term (Decreasing AP)	First term $a_n$ where $a_n < 0$.	Find smallest integer $n$ such that $a_n < 0$.

Additional Information: Properties of AP

Understanding the properties of an Arithmetic Progression is key to solving problems like finding specific terms or the sum of terms.

If $d > 0$, the AP is increasing.
If $d < 0$, the AP is decreasing.
If $d = 0$, the AP is constant.
The sum of the first $n$ terms of an AP is given by $S_n = \frac{n}{2}(2a + (n-1)d)$ or $S_n = \frac{n}{2}(a + a_n)$.
The terms of an AP are equally spaced on the number line.

In this specific problem, since the common difference $d = -\frac{7}{12}$ is negative, the terms are getting smaller. This is why the first term that becomes negative is the largest negative term (closest to zero on the negative side).

Prepp Team · Answer

5

Prepp Team · Answer

7

Prepp Team · Answer

n cannot be determined