The production of a TV sets in a factory increases uniformly by a fixed number every day. It produced 16000 sets in 6th year and 22600 in 9th year
Questions:
1:Find the production during first year
2)Find the production during 8th year
3)In which year the production is 29200
4)Find the difference of the production during 7th and 4th year
Answers
Step-by-step explanation:
a(n) = a + (n-1) d
According to Case I
16000 = a + ( 6-1)d
16000 = a + 5d. -(I)
According to Case II
22600 = a + 8d. -(ii)
By equation (I) and (ii)
a + 5d = 16000
a + 8d = 22600
- - -
_____________________
-3d = -6600
d = 2200
Putting the value of d in eq.(I)
a + 5(2200) = 16000
a = 16000 - 11000
a = 5000
So answer of 1st question in 5000
2. a + (n-1) d = a(n)
5000 + (8-1)2200
5000 + 7 × 2200
5000 + 15400
= 20400
3. 29200 = 5000 + (n-1) 2200
29200 = 5000 + 2200n -2200
29200 = 2800 + 2200n
29200 - 2800 = 2200n
26400 = 2200n
26400/2200 = n
12 = n
4. Production in 7th year =
5000 + 6(2200) = 18800
Production in 4th year =
5000 + 3(2200) = 11600
Difference between them :-
18800 - 11600 = 7200. Ans.