A) 5 : 7
B) 7 : 16
C) 12 : 11
D) None of these
Correct Answer: B
Solution :
[b]: We have, \[\frac{{{S}_{n}}}{S_{n}^{'}}=\frac{(n/2)[2a+(n-1)d]}{(n/2)[2a'+(n-1)d']}=\frac{3n+8}{7n+15},\]where \[{{S}_{n}}\]and \[S{{'}_{n}}\]are sum of n terms of given two A. P. or\[\frac{a+\left( \frac{n-1}{2} \right)d}{a'+\left( \frac{n-1}{2} \right)d'}=\frac{3n+8}{7n+15}\] Choosing (n - 1)/2 = 11 or n = 23 in (i), we get \[\frac{{{T}_{12}}}{T{{'}_{12}}}=\frac{a+11d}{a'+11d'}=\frac{7}{16}\]
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