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 last rungs are [ 2 and(1/2) ]m apart, what is the length of the wood required for the rungs?

 cbse ncert 10th mathematics

 chapter 5 arithmetic progressions, exercise 5.4 optional exercise

 

 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 last rungs are [ 2 and(1/2) ]m apart, what is the length of the wood required for the rungs?

top and last rungs are [ 2 and(1/2) ]m apart means

top and last rungs are [5/2]m apart

 

changing to cm.

top and last rungs are [5/2]*100 =250 cm apart

 

now consecutive rungs are 25cm apart 

so number of rungs =[250/25] +1 = 10+1 =11 rungs

The rungs decrease uniformly in length from 45 cm at the bottom to 25 cm at the top means 

 

counting upwards

 the length of the rungs form an AP  with 

n=11terms

a=45cm 

last term (11th term ) t(11) =25cm

 

To find the total length of wood we find the sum of lengths S(11)

S(n) = (n/2) [a + L ] , L refers to the last term , L = t(11) =25cm

 

S(11) = (11/2) [ 45 + 25 ]

=(11/2)[70] = 11*35

=385 cm

 

 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.

a = 1

The house numbers form an AP with

a=1 and 

kth term =t(k) =k for k =1 to 49 

using S(n) = (n/2) [a + L ] , where 

where L refers to the last term 

 

Sx = (x/2)[1+x]

 

S(49) = (49/2) [1+49] = (49/2)[50]=49*25=1225

 

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

 

by given conditions

 S(x-1) = S(49) - S(x)

[(x-1)/2][x] = 1225 - (x/2)[1+x]

 

multiply each term with 2

(x-1)x = 2450 -x(1+x)

(x^2) - x = 2450 -x - (x^2) 

2(x^2)  -2450 = 0

 dividing by 2

(x^2)  - 1225 = 0

 

(x^2) = 1225

taking square root

x = 35 , because x cannot be negative

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ncert cbse 10th mathematics

chapter 5  arithmetic progressions 

exercise 5.4 optional exercise

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

solution

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

 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 last rungs are [ 2 and(1/2) ]m apart, what is the length of the wood required for the rungs?

solution

 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

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