A) 26
B) 25
C) 23
D) 20
Correct Answer: D
Solution :
Let the number of months be n. Then, \[3250=20+35+50+...\] to n terms It is an Arithmetic sequence with first term, \[a=20\]and common difference, \[d=15\] Applying \[s=\frac{n}{2}[2a+(n-1)d],\] we have \[3250=\frac{n}{2}[2\times 20+(n-1)\times 15]\] \[\Rightarrow \] \[3250=\frac{n}{2}[40+15n-15]\] \[\Rightarrow \] \[6500=n\,(15n+25)\] \[\Rightarrow \] \[15{{n}^{2}}+25n=6500\] \[\Rightarrow \] \[1300=3{{n}^{2}}+5n\] \[\Rightarrow \] \[3{{n}^{2}}+5n-1300=0\] \[\Rightarrow \] \[3{{n}^{2}}+65n-60n-1300=0\] \[\Rightarrow \] \[n(3n+65)-20(3n+65)=0\] \[\Rightarrow \] \[(n-20)(3n+65)=0\] \[\Rightarrow \] \[n=20,\] as n cannot be negative.You need to login to perform this action.
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