Arithmetic Progressions ⇒⇒ Exercise 5.4

Question 1

Which term of the AP : 121, 117, 113,......, is its first negative term?

Solution :

According to the question

Given, First term, a = 121

Common difference, d = 117−121 = −4

The formula for the n th term of an AP is:

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

We need to find the smallest integer value of n for which $ a_n < 0 $

We know that first negative term will be less than zero.

$$ a_n = a + ( n - 1) d < 0 $$

$$ ⇒ 121 + (n - 1)(-4) < 0 $$

$$ ⇒ 121 - 4n + 4 < 0 $$

$$ ⇒ 125 - 4n < 0 $$

$$ ⇒ 125 < 4n $$

$$ ⇒ {125\over 4} < n $$

$$ ⇒ 31.25 < n $$

Since 'n' must be an integer (representing the term number), the smallest integer greater than 31.25 is 32.

Let's check the 32nd term to confirm it is the first negative term:

$$ ⇒ 121 + (32 - 1)(-4) $$

$$ ⇒ 121 + (31)(-4) $$

$$ ⇒ 121 -124 = -3 $$

The 32nd term is -3, which is the first term in the sequence to be negative.

Question 2

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.

Solution :

According to the question

Given, The sum of the 3rd and 7th terms is 6

$$ a_3 + a_7 = 6 $$

$$⇒ (a + 2d) + (a + 6d) = 6 $$

$$ ⇒ 2a + 8d = 6 $$

$$ ⇒ a + 4d = 3 $$

$$ ⇒ a = 3 - 4d .... equation(1) $$

Given, The product of the 3rd and 7th terms is 8:

$$ a_3 × a_7 = 8 $$

$$⇒ (a + 2d) × (a + 6d) = 8 $$

$$ ⇒( (3−4d) + 2d) × ((3−4d) + 6d)= 8 $$

$$ ⇒ (3 − 2d) × (3 + 2d) = 8$$

Using the algebraic identity : $ (x−y)(x+y) = x^2 - y^2 $

$$⇒ 3^2 - (2d)^2 = 8 $$

$$ ⇒ 9 - 4d^2 = 8 $$

$$ ⇒ 1 = 4d^2 $$

$$ ⇒ d^2 = {1\over 4} $$

$$ ⇒ d {\pm}{1\over 2} $$

Since there are two possible values for the common difference, there are two possible APs.

Case 1: Substituting value of d = 1 /2 in equation (1), we get

$$ a = 3 - 4d $$

$$ a = 3 - 2 = 1 $$

We know that , Sum of first n terms

$$ S_n = {n\over 2}[ { 2a + ( n - 1) d } ] $$

Hence, Calculating the Sum of the First 16 Terms

$$ ⇒ S_{16}= {16\over 2} × [ { 2( 1 ) + (16 - 1) ×{{1\over 2}} } ] $$

$$⇒ S_{16}= 8 × [ { 2 + 15 × {1\over 2} } ] $$

$$⇒ S_{16}= 8 × [ { 2 + 7.5 } ] $$

$$ ⇒ S_{16}= 8 × 9.5 $$

$$ ⇒ S_{16}= 76 $$

Case 2: Substituting value of d = $ - {1\over 2} $ in equation (1), we get

$$ a = 3 - 4d $$

$$ a = 3 + 2 = 5 $$

We know that , Sum of first n terms

$$ S_n = {n\over 2}[ { 2a + ( n - 1) d } ] $$

Hence, Calculating the Sum of the First 16 Terms

$$ ⇒ S_{16}= {16\over 2} × [ { 2( 5 ) + (16 - 1) ×{ - {1\over 2}} } ] $$

$$⇒ S_{16}= 8 × [ { 10 + 15 × -{1\over 2} } ] $$

$$⇒ S_{16}= 8 × [ { 10 - 7.5 } ] $$

$$ ⇒ S_{16}= 8 × 2.5 $$

$$ ⇒ S_{16}= 20 $$

Therefore, Two possible sums for the first 16 terms. The two possible sums are 76 and 20.

Question 3

A ladder has rungs 25 cm apart. 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{1\over 2} $ m apart, what is the length of the wood required for the rungs?

Solution :

According to the question

Distance between the consecutive rungs = 25 cm

Distance between the top and bottom rungs = $ 2{1\over 2} $ m = 250cm

Total Number of spaces = Total Distance / Distance between rungs = $ {250 \over 25} $ = 10

Since the number of rungs is always one more than the number of spaces, the total number of rungs is:

Number of rungs = 10 + 1 = 11

The lengths of the rungs form an arithmetic progression (AP), with the length decreasing uniformly. We can find the total length of the wood by summing the lengths of all the rungs. We use the formula for the sum of an AP:

Given, First term, a = 45

last term, l = 25

We know that , Sum of first n terms

$$ S_n = {n\over 2}[ {a + l } ] $$

Hence, Calculating the Sum of the First 11 Terms

$$ ⇒ S_{11}= {11\over 2} × [ 45 + 25 ] $$

$$⇒ S_{11}= {11\over 2} × 70 $$

$$⇒ S_{11}= 11 × 35 $$

$$ ⇒ S_{11}= 385 cm$$

Therefore, The total length of the wood required for the rungs is 385 cm.