A) \[2:3\]
B) \[3:4\]
C) \[4:3\]
D) \[5:6\]
Correct Answer: C
Solution :
Let \[{{S}_{n}}\] and \[S{{'}_{n}}\] be the sums of n terms of two A.P.'s and \[{{T}_{11}}\] and \[T{{'}_{11}}\] be the respective \[{{11}^{th}}\] terms, then \[\frac{{{S}_{n}}}{S{{'}_{n}}}=\frac{\frac{n}{2}[2a+(n-1)d]}{\frac{n}{2}[2a'+(n-1)d']}=\frac{7n+1}{4n+27}\] \[\Rightarrow \] \[\frac{a+\frac{(n-1)}{2}d}{a'+\frac{(n-1)}{2}d'}=\frac{7n+1}{4n+27}\] Now put \[n=21\], we get \[\frac{a+10d}{a'+10d'}=\frac{{{T}_{11}}}{T{{'}_{11}}}=\frac{148}{111}=\frac{4}{3}\]. Note : If ratio of sum of \[n\] terms of two A.P.'s are given in terms of \[n\] and ratio of their \[{{p}^{th}}\] terms are to be found then put \[n=2p-1\]. Here we put \[n=11\times 2-1=21\].
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