Answer : (b) 148 : 111
Let the two A.P.’s be a, a + d, a + 2d, ....... and A, A + D, A + 2D, .......
Given ,
\(\frac{n/2[2a + (n-1)d]}{n/2 [2A + (n-1) D]} = \frac{7n+1}{4n+27}\)
⇒ \(\frac{2a + (n-1)d}{2A + (n-1) D} = \frac{7n+1}{4n+27}\) ... (i)
Now, we have to find the ratio \(\frac{t_{11}}{T_{11}}\) = \(\frac{a+10d}{A+10D}\) = \(\frac{2a+20d}{2A+20D}\)
Putting n = 21 in (i), we get
\(\frac{2a+20d}{2A+20D}\) = \(\frac{7 \times 21 +1}{4 \times 21 +27}\) = \(\frac{147+1}{84+27} = \frac{148}{111}
\)
∴ t11 : T11 = 148 : 111.