A) 14
B) 15
C) 16
D) 17
E) 18
Correct Answer: B
Solution :
Given, AP is 2, 4, 6,... Here first term\[a=2\]and common difference\[d=4-2=2\] \[\because \] \[{{S}_{n}}=\frac{n}{2}[2a+(n-1)d]\] \[\Rightarrow \] \[240=\frac{n}{2}[2\times 2+(n-1)2]\] \[\Rightarrow \] \[480=n4+2n-2]\] \[\Rightarrow \] \[480=2n+2{{n}^{2}}\] \[\Rightarrow \] \[{{n}^{2}}+n-240=0\] \[\Rightarrow \]\[{{n}^{2}}+16n-15n-240=0\] \[\Rightarrow \] \[(n-15)(n+16)=0\] \[\Rightarrow \] \[n=15,n=-16\] \[n=-16\]is not possible. \[\therefore \] \[n=15\]
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