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Q1:
Which term of the AP : 121, 117, 113, . . ., is its first negative term?
[Hint : Find $n$ for $a_n < 0$]
Solution :
Given: An Arithmetic Progression (AP) with terms $121, 117, 113, \dots$
To Find: The value of $n$ such that the $n^{th}$ term ($a_n$) is the first negative term of the sequence.
Step 1: Identify the parameters of the Arithmetic Progression.
The general form of an AP is $a, a+d, a+2d, \dots$ where $a$ is the first term and $d$ is the common difference.
From the given sequence:
First term ($a$) = $121$
Common difference ($d$) = $a_2 - a_1 = 117 - 121 = -4$
Step 2: State the formula for the $n^{th}$ term of an AP.
The formula for the $n^{th}$ term of an AP is given by:
$a_n = a + (n - 1)d$
[Where $a_n$ is the $n^{th}$ term, $a$ is the first term, $n$ is the position of the term, and $d$ is the common difference.]
Step 3: Set up the inequality to find the first negative term.
We are looking for the first term that is less than zero. Therefore, we set $a_n < 0$:
$a + (n - 1)d < 0$
Substitute the known values $a = 121$ and $d = -4$ into the inequality:
$121 + (n - 1)(-4) < 0$
Step 4: Solve the inequality for $n$.
$121 - 4n + 4 < 0$ [Distributing $-4$ into the parentheses]
$125 - 4n < 0$ [Combining like terms $121 + 4 = 125$]
$-4n < -125$ [Subtracting $125$ from both sides]
$4n > 125$ [Multiplying by $-1$ reverses the inequality sign]
$n > \frac{125}{4}$ [Dividing both sides by $4$]
$n > 31.25$
Step 5: Determine the integer value for $n$.
Since $n$ must be a positive integer representing the position of a term in the sequence, and we require the smallest integer $n$ such that $n > 31.25$, we conclude that $n = 32$.
Step 6: Verification (Optional but recommended).
Calculate the $31^{st}$ term: $a_{31} = 121 + (31 - 1)(-4) = 121 + 30(-4) = 121 - 120 = 1$. (This is positive)
Calculate the $32^{nd}$ term: $a_{32} = 121 + (32 - 1)(-4) = 121 + 31(-4) = 121 - 124 = -3$. (This is the first negative term)
Final Answer: The $32^{nd}$ term of the AP is its first negative term.
More Questions from Class 10 Mathematics Arithmetic Progression EXERCISE 5.4
- Q2: The sum of the third and the seventh terms of an AP is 6 and their product is 8. Find the sum of first sixteen terms of the AP.
- Q3: A ladder has rungs 25 cm apart. (see Fig. 5.7). The rungs decrease uniformly in length from 45 cm at the bottom to 25 cm at the top. If the top and the bottom rungs are $2\frac{1}{2}$ m apart, what is the length of the wood required for the rungs? [Hint : Number of rungs = $\frac{250}{25} + 1$]
- Q4: The houses of a row are numbered consecutively from 1 to 49. Show that there is a value of $x$ such that the sum of the numbers of the houses preceding the house numbered $x$ is equal to the sum of the numbers of the houses following it. Find this value of $x$. [Hint : $S_{x-1} = S_{49} - S_x$]
- Q5: A small terrace at a football ground comprises of 15 steps each of which is 50 m long and built of solid concrete. Each step has a rise of $\frac{1}{4}$ m and a tread of $\frac{1}{2}$ m. (see Fig. 5.8). Calculate the total volume of concrete required to build the terrace. [Hint : Volume of concrete required to build the first step = $\frac{1}{4} \times \frac{1}{2} \times 50$ m$^3$]
CBSE Solutions for Class 10 Mathematics Arithmetic Progression
Chapters in CBSE - Class 10 Mathematics
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