a man borrows rs8000 and agrees to repay with total interest of the rs1360 in 12 monthly installments .each installment begin less than preeciding one by rs 40 find amount of first and last installment
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n=12 sn=s12 =8000+1360=9360. d=-40. formula. sn=n/[2a+(n-1)d]. s12=9360(2a 12/2[12-1]×-40 9360=6(2a+(11)×-40 9360=6(2a-440). 6(2a-440)=9360 2a-440=9360/6. 2a-440=1560. 2a=2000. a=1000. formula tn=a+(n-1)d t12 =1000+(12-1)×(-40). =1000+11×(-40). 1000-440. tn. 560
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