Arithmetic Progressions | Exercise 5.4

Question 1

Which term of the AP : 121,117,113,…121, 117, 113, \dots, is its first negative term?

[Hint : Find nn for an<0a_n < 0]

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Solution
Understand the Question
  • For an Arithmetic Progression (AP) with first term aa and common difference dd, the nthn^{\text{th}} term is given by an=a+(n−1)da_n = a + (n - 1)d.
  • To find the first negative term, we set an<0a_n < 0 and solve the inequality for nn.
  • Since the term number nn must be a positive integer, we round up to the smallest integer greater than the calculated value.

Step 1 · Find the First Term and Common Difference

Given AP: 121,117,113,…121, 117, 113, \dots

First term, a=121a = 121

Common difference,

d=117−121=−4\begin{aligned} d &= 117 - 121 \\ &= -4 \end{aligned}

Step 2 · Set Up and Solve the Inequality

The nthn^{\text{th}} term of an AP is: an=a+(n−1)da_n = a + (n - 1)d

For ana_n to be the first negative term, set an<0a_n < 0: a+(n−1)d<0a + (n - 1)d < 0

Substitute a=121a = 121 and d=−4d = -4: 121+(n−1)(−4)<0121 + (n - 1)(-4) < 0

121−4n+4<0125−4n<0\begin{aligned} 121 - 4n + 4 &< 0 \\[0.6em] 125 - 4n &< 0 \end{aligned}

125<4n125 < 4n

1254<n31.25<n\begin{aligned} \dfrac{125}{4} &< n \\[0.8em] 31.25 &< n \end{aligned}

Since nn must be a positive integer, the smallest integer value greater than 31.2531.25 is n=32n = 32.

Answer

32nd32^{\text{nd}} term

Common Mistakes
  • Rounding Down: Mistakenly rounding 31.2531.25 down to 3131. For n=31n = 31, a31=121+30(−4)=1a_{31} = 121 + 30(-4) = 1, which is still positive. The first negative term occurs at n=32n = 32.
  • Inequality Direction: Misinterpreting 31.25<n31.25 < n as n<31.25n < 31.25. Since nn is greater than 31.2531.25, we must choose the next whole integer.

More questions in Exercise 5.4

Q1

Which term of the AP : 121,117,113,…121, 117, 113, \dots, is its first negative term?

[Hint : Find nn for an<0a_n < 0]

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 212 m2\dfrac{1}{2}\text{ m} apart, what is the length of the wood required for the rungs?

[Hint : Number of rungs = 25025+1\dfrac{250}{25} + 1]

Q4

The houses of a row are numbered consecutively from 1 to 49. Show that there is a value of xx such that the sum of the numbers of the houses preceding the house numbered xx is equal to the sum of the numbers of the houses following it. Find this value of xx.

[Hint : Sx−1=S49−SxS_{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 14 m\dfrac{1}{4} \text{ m} and a tread of 12 m\dfrac{1}{2} \text{ 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 = 14×12×50 m3\dfrac{1}{4} \times \dfrac{1}{2} \times 50 \text{ m}^3]

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