A) 3
B) 5
C) 8
D) Both 'a' and 'c'
Correct Answer: D
Solution :
\[{{S}_{n}}=\frac{n}{2}\,[2a+(n-1)d]\] \[\Rightarrow \] \[48=\frac{n}{2}\,[\,2\times 20+(n-1)(-4)]\] \[\Rightarrow \] \[96=40n-4{{n}^{2}}+4n\] \[\Rightarrow \] \[{{n}^{2}}-11n+24=0\] \[\Rightarrow \] \[(n-8)\,(n-3)=0\] \[\therefore \] \[n=8\] or 3
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