A) \[{{a}_{1}}=-10\]
B) \[{{a}_{2}}=-1\]
C) \[{{a}_{3}}=-4\]
D) \[{{a}_{5}}=2\]
Correct Answer: B
Solution :
\[\because \]\[{{a}_{1}}+{{a}_{3}}+{{a}_{5}}=-12\] \[a+(a+2d)+(a+4d)=-12\] \[(\because d>0)\] \[\Rightarrow \] \[a+2d=-4\] ...(i) and \[{{a}_{1}}{{a}_{3}}{{a}_{5}}=80\] \[\Rightarrow \]\[a(a+2d)(a+4d)=80\] \[\Rightarrow \] \[a(-4)(-4-2d+4d)=80\] [from Eq. (i)] \[\Rightarrow \] \[(-4-2d)(-4)(-4-2d+4d)=80\] \[\Rightarrow \] \[4(4+2d)(-4+2d)=80\] \[\Rightarrow \] \[4(4{{d}^{2}}-16)=80\] \[\Rightarrow \] \[4{{d}^{2}}=16+20=36\] \[\Rightarrow \] \[{{d}^{2}}=9\] \[\therefore \] \[d=\pm 3\] Since, AP is increasing, then \[d=+3;a=-10\] \[\therefore \] \[{{a}_{1}}=-10;{{a}_{2}}=-7\] \[{{a}_{3}}=a+2d=-10+6=-4\] \[{{a}_{5}}=a+4d=-10+8=-2\]You need to login to perform this action.
You will be redirected in
3 sec