NCERT Solutions For Class 10 Math Chapter – 5 Exercise – 5.4

NCERT Solutions For Class 10 Math Chapter – 5 Exercise – 5.4

Q1. Which term of the AP : 121 , 117 , 113 ,….., is its first negative term ?

Solution:-

A_n = 121 + (n-1)(-4)

= 121 – 4n + 4

= 125 – 4n

125 – 4n < 0

125 < 4n

N > 125/4

N > 31.25

Q2. The sum of the third and the seventh terms of an AP is 6 and their product is 8. Find the sum of the first sixteen terms of the AP.

Solution:-

A₃ + a₇ = a₆(I)

And

A₃ x a₇ = a₈ (ii)

By the nth term formula,

A_n = a + (n-1)d

Third term a₃ = a + (3-1)d

A₃ a + 2d (iii)

A₇ = a + 6d (iv)

From equation (iii) and (iv) putting in equation (I) we get

A + 2d + a + 6d = 6

2a + 8d = 6

A + 4d = 3

A = 3 – 4d (v)

Again putting the equation (iii) and (iv) in equation (ii) we get

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

Putting the value of a from equation (v) we get

(3-4d+2d) x (3-4d+6d) = 8

4d² = 1

D = 1/2 or -1/2

Now , by putting both the values of d , we get

A = 3 – 4d = 3 – 4(1/2) = 3 – 2 = 1 when d = 1/2

A = 3 – 4d = 3 – 4(-1/2) = 3 + 2 = 5 when d = -1/2

We know the sum of the nth term of AP is;

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

So , when a = 1 d = 1/2

Then , the sum of first 16 terms are;

S₁₆ = 16/2 [ 2 + (16-1) 1/2]

= 8 (2 + 15/2)

= 76

And when a = 5 and d = -1/2

Then , the sum of first 16 terms are;

S₁₆ = 16/2 [ 2.5 + (16-1)(-1/2)]

= 8(5/2)

= 20

Q3. A ladder has rungs 25 cm apart The rungs decreases uniformly in length from 45 cm at the bottom to 25 cm at the top. If the top and the bottom rungs are 5/2 apart , what is the length of the wood required for the rungs ?

Solution:-

Distance between the top rungs and bottom rung of the ladder is = 5/2 x 100cm

= 250 cm

Total number of rungs = 250/25 + 1 = 11

S_n = n/2 (a + l)

S_n = 11/2 (45 + 25)

S_n = 385 cm

Q4. The houses of a row are numbered consecutively from 1 to 49. Show that there is a value of x such that 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:-

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

= (X-1)/2 [2.1 + (X-1)1]

= (X-1)/2 [2 +X – 2]

= X(X-1)/2 (I)

S_49-x = 49/1[2.1+ (49 – 1)1]

= 25(49) – x(x+1)/2 (ii)

X(x-1)/2 = 25(49) – x(x-1)/2

X = 35

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 and a tread of 12 m. Calculate the total volume of concrete required to build the terrace.

Solution:-