A) \[-\,513\]
B) \[-\,171\]
C) \[\frac{511}{3}\]
D) \[171\]
Correct Answer: B
Solution :
[b] |
Let the G.P. be \[a,\text{ }ar,\text{ }a{{r}^{2}},\,a{{r}^{3}},\,......\]and\[a<0\]. |
\[\because \,\,\,\,\,\,\,\,{{a}_{1}}+{{a}_{2}}=4\Rightarrow a\,(1+r)=4\] ...(i) |
\[{{a}_{3}}+{{a}_{4}}=16\Rightarrow a{{r}^{2}}\text{(}1+r\text{)}=16\] ...(ii) |
\[\therefore \] from (i) and (ii), \[r=\pm \,2\]. |
if r = 2, then \[a=\frac{4}{3}\] |
if r = -2, then a = -4. |
\[\therefore \,\,\sum\limits_{i=1}^{9}{{{a}_{i}}}=\frac{a({{r}^{9}}-1)}{r-1}=4\lambda \] |
\[=\frac{-\,4\cdot ({{(-2)}^{9}}-1)}{-\,2-1}=4\lambda \] |
\[\therefore \,\,\lambda \,=-\,171\] |
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