Arithmetic Progression NCERT Exercise 5.4 Class Ten Mathematics

# Arithmetic Progression

## Exercise 5.4

Question: 1. Which term of the AP: 121, 117, 113, . . ., is its first negative term?

Solution: Here; a = 121 and d = -4

The first negative term will be less than zero. Moreover, each term gives 1 as remainder when divided by 4. So the last positive term should be 5. Let us check the value of n for an = 5.

5 = a + (n – 1)d

Or, 121 + (n – 1) (- 4) = 5

Or, 121 – 4n + 4 = 5

Or, 125 – 4n = 5

Or, 4n = 120

Or, n = 30

Thus, a_(31) = 5 – 4 = 1

Hence, 32nd term will be the first negative term.

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: We can solve this problem as follows:

a_3 = a + 2d

a_7 = a + 6d

As per question;

a_3 + a_8 = 6

Or, a + 2d + a + 6d = 6

Or, 2a + 8d = 6

Or, a + 4d = 3

Or, a = 3 – 4d ……. (1)

Similarly,

(a + 2d)(a + 6d) = 8

Or, a^2 + 6ad + 2ad + 12d^2 = 8

Substituting the value of a in equation (2), we get;

(3 – 4d)^2 + 8(3 – 4d)d + 12d^2 = 8

Or, 9 – 24d + 16d^2 + 24d – 32d^2 + 12d^2 = 8

Or, 9 – 4d^2 = 8

Or, 4d^2 = 9 – 8 = 1

Or, 2d = 1

Or, d =1/2

Using the value of d in equation (1), we get;

a = 3 – 4d

Or, a = 3 – 2 = 1

Sum of first 16 terms can be calculated as follows:

S=n/2[2a+(n-1)d]

=(16)/(2)(2xx1+15xx1/2)

=8(2+(15)/(2))

=8xx(19)/(2)=76

Thus, sum of given AP = 76

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 2.5 m apart, what is the length of the wood required for the rungs?

Solution: Number of rungs can be calculated as follows:

2.5 xx 100  cm ÷ 25 + 1

= 2.5 xx 4 + 1

= 10 + 1 = 11

Now, we have; a = 25, a11 = 45 and n = 11

So, d can be calculated as follows:

a_(11) = a + 10d

Or, 45 = 25 + 10d

Or, 10d = 45 – 25 = 20

Or, d = 2

Now, total length of wood shall be equal to the sum of 11 terms;

S=n/2[2a+(n-1)d]

=(11)/(2)(2xx25+10xx2)

=11(25+10)

=11xx35=385 cm

Question: 4. 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.

Solution: Here, a = 1, d = 1 and a49 = 49

As per question;

S_(x-1)=S_(49)-S_x -------(1)

S_x=x/2[2+(x-1)xx1]

=x/2(x+1)

=(x^2+x)/(2)

Similarly,

S_(x-1)=(x-1)/(2)[2+(x-1-1)1]

=(x-1)/(2)(2+(x-2))

=((x-1)x)/(2)

=(x^2-x)/(2)

Similarly,

S_(49)=(49)/(2)(2+48xx1)

=(49)/(2)xx50=1225

After substituting the values of Sx-1, S49 and Sx in equation (1), we get;

S_(x-1)=S_(49)-S_x

Or, (x^2-x)/(2)=1225-(x^2+x)/(2)

Or, (x^2-x)/(x)+(x^2+x)/(2)=1225

Or, (x^2-x+x^2+x)/(2)=1225

Or, (2x^2)/(2)=1225

Or, x^2=1225

Or, x=sqrt(1225)=35

Question 5: 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 ¼ m and a tread of ½ m. Calculate the total volume of concrete required to build the terrace.

Solution: Dimensions in 1st step = 50 m x 0.25 m x 0.5 m

Volume of first step = 6.25 cubic m

Dimensions of second step = 50  m xx 0.5  m xx 0.5  m

Volume of 2nd step = 12.5  cubic  m

Dimensions of 3rd step = 50  m xx 0.75  m xx 0.5  m

Volume of 3rd step = 18.75  cubic  m

Now, we have a = 6.25, d = 6.25 and n = 15

Sum of 15 terms can be calculated as follows:

S=n/2[2a+(n-1)d]

=(15)/(2)(2xx6.25+14xx6.25)

=(15)/(2)xx100=750

Volume of concrete = 750 cubic m