# Class10 NCRT Arithmetic Progressions Exercise – 5.4 (Optional)* pdf || UP Board

# __Arithmetic
Progressions__

__Exercise
– 5.4 (Optional____)__^{*}

__Arithmetic Progressions__

__Exercise – 5.4 (Optional__

__)__

^{*}

**1. Which term of the A.P.: 121, 117, 113, …**** ****is its first negative term****?**

**[****Hint****: ****Find n for a _{n}**

**< 0**

**]**

Here a = 121 and d = 117 – 121 = -4

Let, nth term of AP is the first negative term.

a_{n} < 0

a + (n – 1)d < 0

121 + (n – 1)(-4) < 0

121 – 4n + 4 < 0

125 < 4n

n > 125/4

n > 31.25

n = 32

Hence, 32^{nd} term of this AP is its first
negative term.

**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.**** **

Let, the first term = a and common difference = d

The sum of the third and the seventh terms of an AP is
6, Therefore

a_{3} + a_{7} = 6

a + (3 – 1)d + a + (7 – 1)d = 6

a + 2d + a + 6d = 6

2a + 8d = 6

a + 4d = 3

a = 3 – 4d …………….(1)

The product of the third and the seventh terms of an
AP is 8, Therefore

(a_{3})(a_{7}) = 8

(a + (3 – 1)d)(a + (7 – 1)d) = 8

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

Substituting the value of ‘a’ from Equation (1),

(3 – 2d)(3 + 2d) = 8

3^{2} – 4d^{2} = 8

4d^{2} = 1

d = ± ½

If d = ½ Substituting
the value of ‘d’ from Equation (1),

a = 3 – 4(½) = 1

the sum of first sixteen terms of the AP,

S_{16} = 16/2 [2a + (16 – 1)d] = 8[2(1) +
15(½)] = 76

If d = - ½ Substituting the value of ‘d’ from Equation (1),

a = 3 – 4(-½) = 5

the sum of first sixteen terms of the AP,

S_{16} = 16/2 [2a + (16 – 1)d] = 8[2(5) +
15(-½)] = 20

Hence, the sum of first sixteen terms of this AP is 20 or 76

**3. ****A
ladder has rungs 25 cm apart. (See Figure 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
to 25 cm at the top. If the top and the bottom rungs are**** ****2½**** ****m apart****, ****what is the length of the wood
required for the rungs****?**

**[****Hint:
Number of rungs**** ****= 250/25 + 1****]**

If the distance between the upper and lower rungs is 2½ m and the consecutive rungs are at a distance of 25 cm from each other. Therefore,

Number of rungs = 250/25 + 1 = 10 + 1 = 11

The length of the rungs is increasing from 25 cm to 45 cm in the Arithmetic Progression(AP), in which the
first term a = 25
and the last term a_{11} = 45. Suppose, The common difference of this AP is d.

Therefore, a_{11} = 45

a + (11 – 1)d = 45

25 + 10d = 45

d = 20/10 = 2

the length of the wood required for the rungs = S_{11}

= 11/2[2a + (11 – 1)d] = 11/2[2(25) + 10(2)] = 11 × 35
= 385 cm

Hence, 385 cm length of wood will be required to make the rungs.

**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.**** **

**[****Hint****: ****S _{x – 1} = S_{49}
– S_{x}**

**]**

Here a = 1 and d = 1

Sum of nth terms of AP,

S_{n} = n/2 [2a + (n – 1)d]

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.
Therefore,

S_{x – 1} = S_{49} – S_{x}

(x – 1)/2 [2a + (x – 1 – 1)d] = 49/2 [2a + (49 – 1)d]
– x/2 [2a + (x – 1)d]

(x – 1)/2 [2(1) + (x – 2)1] = 49/2 [2(1) + 48(1)] –
x/2 [2(1) + (x – 1)(1)]

(x – 1)[x] = 49[50] – x[x + 1]

x^{2} – x = 2450 – x^{2} – x

2x^{2} = 2450

x^{2} = 1225

x = 35

Hence, the value of x is 35.

**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 (see figure**** ****5.8). Calculate the total volume
of concrete required to build the terrace.**** **

**[Hint:
Volume of concrete required to build the first step = ¼ × ½ × 50 m ^{3}]**

The volume of concrete used to build the first step = ¼ × ½ × 50 = ¼ × 25 m^{3}

The volume of concrete used to build the second step = 2/4 × ½ × 50 = 2/4 × 25 m^{3}

The volume of concrete used to build the third step = ¾ × ½ × 50 = ¾ × 25 m^{3}

The volume of steps is increasing in AP, in which the
first term a =
¼ × 25 and
the last term a_{15} = 15/4 × 25. Common difference of this AP d = 2/4 × 25 – ¼ × 25 = ¼ × 25

Therefore, the total volume of concrete required to build the
terrace = S_{15}

= 15/2[2(¼ × 25) + (15 – 1)( ¼ × 25)]

= 15/2 [25/2 + 175/2] = 15/2 × 200/2 = 750 m^{3}

Hence, the total volume of concrete required to build the
terrace is 750 m^{3 }

Download Class10 NCRT Arithmetic Progressions Exercise – 5.4 (Optional)* pdf

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