Question

QUESTION OF ARITHEMATIC PROGRESSION
Please all mighty brains

If there are (2n+1) terms in A. P. Then rpove that the ratio of the sum of odd terms and the sum of even terms is (n+1):n.

Answer only if you are sure for solution

Answer 1

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

I think this will help U DEAR

Answer 2

Let a be the first term and d the common difference of the A.P. Also let S1 be the sum of odd terms of A.P. having (2n + 1) terms

$s _{1} = a1 + a3 + a5 + .........a _{2n + 1}$

$s_{1} = \frac{n + 1}{2} (a1 + a_{2n + 1})$

$s_{1} = \frac{n + 1}{2} (a + a \: (2n + 1 - 1) d \: )$

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

Similarly, if S2 denotes the sum of even

terms, then

$s _{2} = \frac{n}{2} (2a + 2nd) = n(a + nd)$

$\frac{s _{1} }{s _{2} } = \frac{(n + 1)(a + nd)}{(n)(a + nd)} = \frac{n + 1}{n}$

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