A) \[\frac{7}{3}a\]
B) \[a\]
C) \[\frac{2}{3}a\]
D) \[5a\]
Correct Answer: B
Solution :
\[b=ar\] \[c=a{{r}^{2}}\] 3a, 7b and 15 c are in A.P. \[\Rightarrow 14b=3a+15c\] \[\Rightarrow 14(ar)=3a+15a{{r}^{2}}\] \[\Rightarrow 14r=3+15{{r}^{2}}\] \[\Rightarrow 15{{r}^{2}}-14r+3=0\]\[\Rightarrow (3r-1)(5r-3)=0\]\[r=\frac{1}{3},\frac{3}{5}.\] Only acceptable value is \[r=\frac{1}{3},\]because\[r\in \left( 0,\frac{1}{2} \right]\] \[\therefore \]\[c.d=7b-3a=7ar-3a=\frac{7}{3}a-3a=-\frac{2}{3}a\] \[\therefore \]\[{{4}^{th}}\,term=15c-\frac{2}{3}a=\frac{15}{9}a-\frac{2}{3}a=a\]You need to login to perform this action.
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