Question

If S1 is the sum of an arithmetic progression of 'n' odd number of terms and S2 the sum of the terms of the series in odd places, then S1/S2 =
Who don't know please don't give answer lol!!​

Answer 1

Answer:

Let n=2m+1

$Then, S1 = \frac{n}{2} (2a1 + (n - 1)d) = \frac{2m + 1}{2} \\ (2a1 +2md)$

$=(2m+1)(a1+md)=n(a+2n−1d)$

$</p><p>S2=a1+a3+a5+...+a2m+1</p><p>$

$</p><p>=21(2n+1)[2a1+(2n+1−1)2d] \\ </p><p></p><p></p><p>=(2n+1)[a1+(2n−1−1)d] \\ </p><p></p><p></p><p>∴S2S1=(2n+1)n=n+12n \\ </p><p></p><p></p><p>$