The middle term(s) of the series \[2+4+6+...+198\]is [SSC (10+2) 20011] |
A) 98
B) 96
C) 84
D) 100
Correct Answer: D
Solution :
It is an arithmetic series. |
where, \[a=2,\]\[{{T}_{n}}=198,\] |
\[d=\]common difference \[=2\] |
Number of terms \[=\,\,n\] |
\[\therefore \] \[{{T}_{n}}=a+(n-1)\,d\] |
\[\Rightarrow \] \[198=2+\,\,(n-1)\,2\] |
\[\Rightarrow \]\[2\,\,(n-1)=198-2=196\] |
\[\Rightarrow \]\[n-1\frac{196}{2}=98\]\[\Rightarrow \]\[n=99\] |
\[\therefore \]Middle term\[=\frac{n+1}{2}=\frac{99+1}{2}=\]50th term |
\[\therefore \] \[{{T}_{50}}=2+(50-1)\,2\] |
\[=2+98=100\] |
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