Let the two arithmetic series be a, a + d, a + 2d ..... and A, A + D, A + 2D, .... .
Given that,
\(\frac{n/2[2a+(n-1)d]}{n/2[2A+(n-1) D]}\) = \(\frac{2n+1}{2n-1}\)
\(\frac{2a+(n-1)d}{2A+(n-1) D}\) = \(\frac{2n+1}{2n-1}\) ...(i)
Ratio of the 10th terms of these series = \(\frac{t_{10}}{T_{10}}\) = \(\frac{a+9d}{A+9D}\) = \(\frac{2a+18d}{2A+18D}\)
∴ Putting n = 19 in (i), we have \(\frac{t_{10}}{T_{10}}\) = \(\frac{2a+18d}{2A+18D}\)
= \(\frac{2 \times 19+1}{2\times 19 - 1}\) = \(\frac{39}{37}\).
