A) Rs.4900
B) Rs.5400
C) Rs.3500
D) Rs.4500
Correct Answer: A
Solution :
1st instalment =Rs. 1000 2nd instalment \[=Rs.\,1000+Rs.\,100=Rs.\,1100\] 3rd instalment \[=Rs.1100+Rs.100=Rs.1200\] and so on Let number of instalments = n \[\therefore \] \[1000+1100+1200+...\]up ton terms =118000 \[\Rightarrow \] \[\frac{n}{2}\,\left[ 2\times 1000+\left( n-1 \right)100 \right]=118000\] \[\Rightarrow \] \[100{{n}^{2}}+1900n-236000=0\] \[\Rightarrow \]\[{{n}^{2}}+19n-2360=0\]\[\Rightarrow \] \[(n+59)\,(n-40)=0\] \[\Rightarrow \] \[n=40\,\,(\therefore n\ne -59)\] \[\therefore \] Total no. of instalments = 40 Now, last instalment = 40th instalment \[\therefore \] \[{{a}_{40}}=a+39d=Rs.4900\]
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