A) 1700
B) 1650
C) 3300
D) 3400
E) 3500
Correct Answer: B
Solution :
Let the first term of an AP is a and common difference is d. \[\therefore \] \[{{T}_{12}}=a+(12-1)d\] \[=a+11d\] and \[{{T}_{22}}=a+(22-1)d\] \[=a+21d\] Since, \[S={{T}_{12}}+{{T}_{22}}\] \[\therefore \] \[100=a+11d+a+21d\] \[100=2(a+16d)\] \[\Rightarrow \] \[a+16d=50\] ?.. (i) Now, \[{{S}_{33}}=\frac{33}{2}[2a+(33-1)d]\] \[=\frac{33}{2}(2a+32d)\] \[=33(a+16d)\] \[=33\times 50\] [from Eq. (i)] \[=1650\]You need to login to perform this action.
You will be redirected in
3 sec