Q. 150 workers were engaged to finish a job in a certain number of days. 4 workers dropped out on second day, 4 more workers dropped out on the third day and so on. It took 8 more days to finish the work. Find the number of days in which the work was completed.

Sol: Suppose the work is completed in n days.

Since 4 workers went away on every day except the first day.

∴ Total number of worker who worked all the n days is the sum of n terms of A.P. with first term 150 and common difference – 4.

Total number of worker who worked all the n days = n/2[2 x 150 + (n-1) x -4 ] = n (152 – 2n)

If the workers would not have went away, then the work would have finished in (n – 8) days with 150 workers working on every day.

∴ Total number of workers who would have worked all n days = 150 (n – 8)

∴ n (152 – 2n) = 150 (n – 8) ⇒ 152n – 2n2 = 150n – 1200 ⇒ 2n2 – 2n – 1200 = 0

⇒ n2 – n – 600 = 0 ⇒ n2 – 25n + 24n – 600 = 0 ⇒ n(n – 25) + 24 (n + 25) = 0

⇒ (n – 25) (n + 24) = 0 ⇒ n – 25 = 0 or n + 24 = 0 ⇒ n = 25 or n = – 24

⇒ n = 25 ( Number of days cannot be negative)

Thus, the work is completed in 25 days.

Q. Divide 32 into 4 parts which are in A.P. such that the product the product of means is 7:15.

Let four numbers in A.P. be

(

⇒ 4

⇒

Product of first and fourth term(extremes) = (

Product of second and third term (means) = (

Product of extremes: product of means = 7:15 (Given)

( 64 – 9

Sol: Suppose the work is completed in n days.

Since 4 workers went away on every day except the first day.

∴ Total number of worker who worked all the n days is the sum of n terms of A.P. with first term 150 and common difference – 4.

Total number of worker who worked all the n days = n/2[2 x 150 + (n-1) x -4 ] = n (152 – 2n)

If the workers would not have went away, then the work would have finished in (n – 8) days with 150 workers working on every day.

∴ Total number of workers who would have worked all n days = 150 (n – 8)

∴ n (152 – 2n) = 150 (n – 8) ⇒ 152n – 2n2 = 150n – 1200 ⇒ 2n2 – 2n – 1200 = 0

⇒ n2 – n – 600 = 0 ⇒ n2 – 25n + 24n – 600 = 0 ⇒ n(n – 25) + 24 (n + 25) = 0

⇒ (n – 25) (n + 24) = 0 ⇒ n – 25 = 0 or n + 24 = 0 ⇒ n = 25 or n = – 24

⇒ n = 25 ( Number of days cannot be negative)

Thus, the work is completed in 25 days.

Q. Divide 32 into 4 parts which are in A.P. such that the product the product of means is 7:15.

Let four numbers in A.P. be

*a*– 3*d*,*a*–*d*,*a*+*d*and*a*+ 3*d.*(

*a*– 2*d*) + (*a*–*d*) + (*a*+*d*) + (*a*+ 2*d*) = 32⇒ 4

*a*= 32⇒

*a*= 8Product of first and fourth term(extremes) = (

*a*– 3*d*) (*a*+ 3*d*) = (8 – 3*d*) (8 +32*d*) = 64 – 9*d*2Product of second and third term (means) = (

*a*–*d*) (*a*+*d*) = (8 –*d*) (8 +*d*) = 64 –*d*2Product of extremes: product of means = 7:15 (Given)

( 64 – 9

*d*2) / (64 –*d*2) = 7 : 15