A) \[11\]
B) \[19\]
C) \[17\]
D) \[18\]
Correct Answer: D
Solution :
Let \[{{A}_{1}},{{A}_{2}},{{A}_{3}}\] and and \[{{A}_{4}}\]are inserted as arithmetic mean between \[-10\] and \[25.\] Then, \[-10,\,{{A}_{1}},{{A}_{2}},{{A}_{3}},{{A}_{4}}\,\,25\] are in AP. Now, \[{{A}_{r}}=a+\frac{r(b-a)}{r+1}\] \[\therefore \] \[{{A}_{u}}=-10+\frac{4(35)}{5}\] \[=-10+28\] \[=18\] \[\therefore \] Fifth term is 18. Alternate \[l={{T}_{n}}=a+(n-1)\,\,d\] \[25=-10+(6-1)\,d\] \[35=5d\] \[\Rightarrow \] \[d=7\] Then, \[{{T}_{5}}=a+4d\] \[=-10+4(7)=-10+28\] \[=18\]
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