A manufacture of TV sets produced 600 seats in the third year 700 sets in the seventh year Assuming that the production increase uniformly by a fixed number even year Find (1) the production in the first (2) the production in the 10th year (3) the total production in the first 7 year
Answers
Answer:
Let 'a' be the first term and 'd' be the common difference of the A.P formed
a
3
=a+2d=600.................(i)
And a
7
=a+6d=700......(ii)
Solving equation (i) and (ii)
a=550 and d=5
(i) Production in the first year is 550 TV sets.
(ii) Production in the 10th year = a
10
=a+9d
=550+9(25)
=550+225
=775
(iii) Total production in 7 years =
S
7
=
2
7
[2(550)+(7−1)25]
=4375.
Thus, the total production in 7 years if of 7375 TV sets
Answer:
it's an A.P.
as the production increases uniformly with fixed amount
we know
A.P.(nth term of a.p.)=a+(n-1)d
we have 3th term=600 and 7th term =700
so 600=a+2d .........(1)
and 700=a+6d.........(2)
on subtracting (1)from(2)
we get
100=4d
i.e. d=25
now put the value of d in above the(1)
we get
600=a+2×25
600=a+50
I.e. a=550
solution 1: production in 1st year =550
solution 2: a.p.(10th) = a+9d
put the value of a and d
we get
a.p.(10th) = 550+9×25
=775
solution 3: sum of first 7 a.p.
I.e. S (sum)=n/2 [2a+(n-1)d]
=7/2 [2×550+6×25]
=7/2 [1100+150]
=7/2 (1250)
=7×625
=4375