A) \[53:155\]
B) \[27:87\]
C) \[29:83\]
D) \[31:89\]
Correct Answer: A
Solution :
| Sum of an A.P. is given by \[{{S}_{n}}=\frac{n}{2}[2a+(n-1)d]\] |
| where 'a' is the first term and 'd' is the common difference of A.P. |
| Let \[{{S}_{{{n}_{1}}}}\] be the sum of n terms of \[{{I}^{st}}\] A.P. |
| and \[{{S}_{{{n}_{2}}}}\] be the sum of n terms of \[I{{I}^{nd}}\] A.P. |
| Given that the sum of n terms of two arithmetic series is in the ratio \[2n+3:6n+5\] |
| \[\Rightarrow \,\,\,\frac{{{S}_{{{n}_{1}}}}}{{{S}_{{{n}_{2}}}}}=\frac{2n+3}{6n+5}\] ?(i) |
| \[\Rightarrow \,\,\,{{S}_{{{n}_{1}}}}=\frac{n}{2}\left[ 2{{a}_{1}}+(n-1){{d}_{1}} \right]=2n+3\] and \[{{S}_{{{n}_{2}}}}=\frac{n}{2}\left[ 2{{a}_{2}}+(n-1){{d}_{2}} \right]=6n+5\] |
| From Eq. (i), we get \[\frac{{{S}_{{{n}_{1}}}}}{{{S}_{{{n}_{2}}}}}=\frac{\frac{n}{2}[2{{a}_{1}}+(n-1){{d}_{1}}]}{\frac{n}{2}[2{{a}_{2}}+(n-1){{d}_{2}}]}=\frac{2n+3}{6n+5}\] |
| \[\Rightarrow \,\,\frac{2{{a}_{1}}+(n-1){{d}_{1}}}{2{{a}_{2}}+(n-1){{d}_{2}}}=\frac{2n+3}{6n+5}\] |
| For \[a=13,\,n=2a-1=2\times 13-1=25\] |
| \[\therefore \,\,\frac{2{{a}_{1}}+(25-1){{d}_{1}}}{2{{a}_{2}}+(25-1){{d}_{2}}}=\frac{53}{155}\Rightarrow \frac{{{a}_{1}}+12{{d}_{1}}}{{{a}_{2}}+12{{d}_{2}}}=\frac{53}{155}\] |
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