Two men X and Y started working for a certain company at similar jobs on January 1, 1950. X asked for an initial salary of Rs. 300 with an annual increment of Rs. 30. Y asked for n initial salary of Rs. 200 with a rise of Rs. 15 every six months. Assume that the arrangements remained unaltered till December 31, 1959. Salary is paid on the last day of the month. What is the total amount paid to them as salary during the period?

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Satt Academy · Accepted Answer

Initial annual salary of X = 12

× 300 = 3600 Tk.Annual increment of X = 12 × 30 = 360 Tk.Total salary of X for 10 years will be found using arithmetic progression,with initial values 3,600 Tk. and difference 360 Tk. for 10 years.So, here n = 10; a = 3600; d = 360 ∴Total salary of X = n2{2a+(n-1)d}=102{(2×3600)+(10-1)360}=5{(7200+(9 ×360)}=5(7200+3240)=5×10440=52200 Tk.Initial half-yearly of Y = 6 × 200 = 1, 200Tkand half-yearly salary of Y = 15 × 6 = 90TkTotal salary of Y for 10 years or 20 half-yearly period will be found using arithmetic progressionwith initial value Tk. 1,200 and difference Tk. 90 for 20 half-year periodSo, here n = 20 a = 1, 200 d = 90 ∴Total salary of Y = n2{2a+(n-1)d}=202{2×1200)+(20-1)20} =10{(2400+(19×90)}=10(2400+1710)=10×4110=41100 Tk.

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