Misc 32 - 150 workers were engaged to finish a job in - Finding sum from nth number

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Misc 32 150 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 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. Let total work = 1 and let total work be completed in = n days Work done in 1 day = (๐‘‡๐‘œ๐‘ก๐‘Ž๐‘™ ๐‘ค๐‘œ๐‘Ÿ๐‘˜)/(๐‘๐‘ข๐‘š๐‘๐‘’๐‘Ÿ ๐‘œ๐‘“ ๐‘‘๐‘Ž๐‘ฆ๐‘  ๐‘ก๐‘œ ๐‘๐‘œ๐‘š๐‘๐‘™๐‘’๐‘ก๐‘’ ๐‘ค๐‘œ๐‘Ÿ๐‘˜) = 1/๐‘› This is the work done by 150 workers Work done by 1 worker in one day = 1/150๐‘› Given that In this manner it took 8 more days to finish the work i.e. work finished in (n + 8) days Therefore, 150/150๐‘› + 146/150๐‘› + 144/150๐‘› + โ€ฆ + to (n + 8)term = 1 1/150๐‘› [150 + 146 + 142 + โ€ฆ to (n + 8)term]= 1 150 + 146 + 142 + โ€ฆ + to (n + 8)terms = 150n 150 + 146 + 142 + โ€ฆ + to (n + 8) terms this is an AP, where first term (a) = 150 Common difference = 146 โ€“ 150 = -4 We know that sum of n terms of AP Sn = ๐‘›/2[2a + (n โ€“ 1)d] Putting n = n + 8 , a = 150 , d = -4 Sn + 8 = (๐‘›+8)/2[2(150) + (n + 8 โ€“ 1)(-4)] = (๐‘›+8)/2 [300 + (n + 7)(-4)] = (๐‘›+8)/2 [300 โ€“ 4 (n + 7)] = (2(๐‘›+8))/2[150 โ€“ 2(n + 7)] = (n + 8)[150 โ€“ 2(n + 7)] = (n + 8)[150 โ€“ 2n โ€“ 14] = n(150 โ€“ 2n โ€“ 14) + 1200 โ€“ 16n โ€“ 112 = 150n โ€“ 2n2 โ€“ 14n -16n + 1200 โ€“ 112 = โ€“ 2n2 + 150n โ€“ 14n โ€“ 16n + 1200 โ€“ 112 = โ€“ 2n2 + 120n + 1088 Hence, 150 +146 +142 + โ€ฆ to (n + 8)term = -2n2 + 120n + 1088 Also, we know that 150 +146 +142 + โ€ฆ to (n + 8)term = 150n โ€“ 2n2 + 120n + 1088 = 150n โ€“ 2n2 + 120n โ€“ 150n + 1088 = 0 โ€“ 2n2 โ€“ 30n + 1088 = 0 2n2 + 30n โ€“ 1088 = 0 n2 + 15n โ€“ 544 = 0 n2 + 32n โ€“ 17n โ€“ 544 = 0 n (n + 32) โ€“ 17(n + 32) = 0 (n โ€“ 17)(n + 32) = 0 Since n cannot be negative , n = โ€“ 32 is not possible Hence n = 17 Work was completed in n + 8 days i.e. 17 + 8 = 25 days

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