| 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 |
| \[\therefore \] \[{{T}_{50}}=2+(50-1)\,2\] |
| \[=2+98=100\] |
You need to login to perform this action.
You will be redirected in
3 sec
