A) 100
B) 200
C) 150
D) 250
Correct Answer: B
Solution :
Series, \[\Rightarrow \] \[a(1-r)=12r\] and let number of terms is \[n\] then sum of A.P.\[{{a}_{1}}+{{a}_{5}}+{{a}_{10}}+{{a}_{15}}+{{a}_{20}}+{{a}_{24}}=225\] \[\Rightarrow \]\[60100=\frac{n}{2}\left\{ 2\times 2+(n-1)3 \right\}\]\[\Rightarrow \]\[120200=n(3n+1)\] \[\Rightarrow \]\[3{{n}^{2}}+n-120200=0\]\[\Rightarrow \]\[(n-200)(3n+601)=0\] Hence\[n=200\].
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