A man Borrows Rs. 8000 and agrees to repay with a total interest of Rs. 1360 in 12 monthly instalments. Each instalment being less than the preceding one by Rs. 40. Find the amount of the first and last instalments.

Answer

The first instalment is Rs. 1000 and the last instalment is Rs.560

Explanation

The amount repaid `=Rs.8000+Rs.1360=Rs.9360` The number of instalments = 12
`thereforen=12,S_(n)=S_(12)=9360`
Each instalments is Rs. 40 less than the preceding one.
`therefored= 40.` This is an A.P.
`S_(n)=(n)/(2)[2a+(n-1)d]` ...(Formula)
`thereforeS_(12)=9360=(12)/(2)[2a+(12-1)xx(-40)` ....(Substituting the values)
`therefore9360=6[2a+(11)xx(-40)]`
`therefore9360=6(2a-440)`
`therefore6(2a-440)=9360`
`therefore2a-440=(93600)/(60)` ...(Dividing both the sides by 6)
`therefore2a-440=1560`
`therefore2a=1560 +440" " therefore2a=2000 " "thereforea=1000`
The last instalment `t_(n)`
`t_(n)=a+(n-1)d` ...(Formula)
`thereforet_(12)=1000+(12-1)xx(-40)`
`=1000+11xx(-40)`
`=1000-440`
`thereforet_(n)=560`