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Sum of N Terms of an AP

In an A.P. if...

Question

In an A.P. if $S_{1} = T_{1} + T_{2} + T_{3} + . . . + T_{n}$ (n odd), $S_{2} = T_{2} + T_{4} + T_{6} + . . . + T_{n - 1}$ , then find the value of $S_{1} / S_{2}$ in terms of n

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Solution

$S_{n} = \frac{n}{2} [2 a + (n - 1) d]$

$S_{1} = \frac{n}{2} [2 a + (n - 1) d]$ ...(1)

$S_{2} = \frac{\frac{n - 1}{2}}{2} {2 (a + d) + [\frac{(n - 1)}{2} - 1] 2 d}$

$S_{2} = \frac{(n - 1)}{4} [2 a + 2 d + (n - 1) d - 2 d]$

$S_{2} = \frac{(n - 1)}{4} [2 a + (n - 1) d]$ ...(2)

$(1) \div (2)$ gives

$\frac{S_{1}}{S_{2}} = \frac{\frac{n}{2} [2 a + (n - 1) d]}{\frac{(n - 1)}{4} [2 a + (n - 1) d]}$

$\frac{S_{1}}{S_{2}} = \frac{n}{2} \times \frac{4}{(n - 1)}$

$\frac{S_{1}}{S_{2}} = \frac{2 n}{n - 1}$

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