A) 7
B) 21
C) 28
D) 42
Correct Answer: C
Solution :
According to Question \[\Rightarrow \]\[\frac{{{S}_{5}}}{S{{'}_{5}}}=49\]{here, \[{{S}_{5}}=\] Sum of first 5 terms and \[{{S}_{5}}=\]Sum of their reciprocals) \[\Rightarrow \frac{\frac{a({{r}^{5}}-1)}{(r-1)}}{\frac{{{a}^{-1}}({{r}^{-5}}-1)}{({{r}^{-1}}-1)}}=49\] \[\Rightarrow \frac{a({{r}^{5}}-1)\times ({{r}^{-1}}-1)}{{{a}^{-1}}({{r}^{-5}}-1)\times (r-1)}=49\] or\[\frac{{{a}^{2}}(\cancel{1-{{r}^{5}}})\times (\cancel{1-r})\times {{r}^{5}}}{(\cancel{1-{{r}^{5}}})\times (\cancel{1-r})\times r}=49\] \[\Rightarrow {{a}^{2}}{{r}^{4}}=49\Rightarrow {{a}^{2}}{{r}^{4}}={{7}^{2}}\]\[\Rightarrow \]\[\]?.(1) Also, given, \[{{S}_{1}}+{{S}_{3}}=35\] \[a+a{{r}^{2}}=35\] ...(2) Now substituting the value of eq. (1) in eq. (2) a + 7 = 35 \[\]
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