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Jan 28, 2025 · Question A manufacturer of TV sets produced 600 sets in the third year and 700 sets in the seventh year. Assuming that the production increases uniformly by a fixed number every year, find: i) the production in the 1st year ii) the production in the 10th year iii) the total production in first 7 years Asked Jan 28 at 00:23 Helpful Report.
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Art limited in composition to the dimensions of depth and height is called 2D art. This includes paintings, drawings and photographs and excludes three-dimensional forms such as sc.
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Production in the third year, P3=P1+2d=600 Production in the seventh year, P7=P1+6d=700Now, we can solve for ddd by subtracting the two equations:P7−P3= (P1+6d)− (P1+2d)=700−6004d=100d=100/4 = 25So, the difference between the productions of two consecutive years is 25.
Feb 25, 2017 · Let the number of sets produced in 1st year be 'a' and 'd' be the increase in production every year. We are given,a+2d= 600 ----- (1) anda+6d= 700 ----- (2) subtracting equation (1) from (2), we get 4d= 100 ord= 25 Substituting d=25 in equation (1), we get a= 550 (a) Production in the first year =a= 550 (b) Production in 10th year = a+ 9d = 550+9 x25 =775 (c) Total production in first 7 years.
We know that: a1 + 2d = 500 (production in the third year is 500) a1 + 6d = 600 (production in the seventh year is 600) From the first equation, we can solve for d: d = (500 - a1)/2 Substituting this value of d in the second equation, we get: a1 + 3 (500 - a1)/2 = 600 Simplifying this equation, we get: a1 = 450 So we know that: a1 = 450 d.
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Disclaimer: Artikel ini dibuat untuk tujuan informasi dan hiburan semata. Mar 14, 2021 · Answer: i) Since the production increases uniformly by a fixed number every year, the number of TV sets manufactured in 1st, 2nd, 3rd, . . ., years will form an AP. Let us denote the number of TV sets manufactured in the nth year by an. Then, a3 = 600 and a7 = 700 or, a + 2d = 600 a + 6d = 700 Solving these equations, we get d = 25 and a = 550. Therefore, production of TV sets in the first.