Question

If there are (2n+1) terms in an AP, then prove that the ratio of the sum of the odd terms and the sum of the even terms is (n+1):n.​

Answer 1

Answer:

There are (2n +1) terms in an arithmetic series.

Let a , a + d , a + 2d , a + 3d , a + 4d ....... + a + (2n +1-1)d

Here, odd terms are : a , a + 2d , a + 4d , a + 6d , ...... a + 2nd

Let number of odd terms = r and common difference = 2d

Tr = a + (r - 1)2d

a + 2nd = a + (r - 1)2d

2nd/2d = r - 1

r = n + 1

Hence, there are (n +1) odd terms

Now, sum of odd terms = (n+1)/2[a + a + 2nd] [∵ Sn = n/2[first term + nth term]]

= (n + 1)(a + nd) -----(1)

Similarly, even terms are a +d , a + 3d , a + 5d , ...... a + (2n -1)d

Number of even terms = 2n+1 - (n +1) = n

Now, sum of even terms = n/2[a + d + a + (2n-1)d]

= n(a + d)

Now, sum of odd terms/sum of even terms = (n+1)(a+d)/n(a+d)

= (n + 1)/n