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Task 2 A small town would like to install 9 renewable lighting poles covering a distance of 2900 m. The first pole will be at 170 m. a. Determine the values of each pole location correct to the nearest whole number using arithmetic progression. an eas+(n-1).d. Here, first pole is at 170m a = 170 (the first term of arithmetic progherion) The compon difference is d = following table gives the distance of each pole rounded to whole number, 2900 -170 8 - 2730 -341.25 8 Pole No. Distance (Rounded) 1 170 170 2 511.25 511 3 852.5 853 4 1193.75 1194 5 1535 1535 1876.25 1876 7 2217.5 2218 8 2558.75 2559 9 2900 2900 b. Determine the values of each pole location correct to the nearest whole number using geometric progression. here, an=arn where a is the first termice a-170 pm So, ag =2900 > 2900=170.784-170x8*470) Va The following table gives the distance of each pol rounded to Whole number: 9 Pole No. Distance (Rounded) 1 170 170 2 3 4 5 6 7 242.3497 345.4906 492.5268 702.1396 1000.961 1426.956 242 345 493 702 1001 1427 8 2034.25 2034 2900 2900 c. If a second town would like to implement the combination of the two methods shown above such that poles are switched between arithmetic and geometric progressions. The first pole will be installed at 150 m and the pole number 25 will be installed at 1350 m. Then poles will be installed based on a geometric model such that it will cover 3.5 km * representing the remaining distance. The ratio between two poles is 1.026. Determine the location for pole number 10 of the new geometric progression model and the approximate total number of poles that are needed for this town. Comment on these results. * Antonow shlemena Que to ambiquity of the problem statement, we consider the anthmetico geometric serico: (As the part may be designed a non-mathematical instructor). a, (ard)r, atrd, 82 Given 25th term at 1350m with a = 150 m, 1026 we have, (1587 24xD) (1.02634 1350 => d= 24.129866995163616175783193586891 m The distance of 10th pole in this case = (150+90) r = 462.5855 = 463 m (rounded value) Number of poles in 3.5 km is given by: 150 + (n-1) d) rn-1 = 3500 na 43.5 44 poles. (Rounded volue) - (1- ) )