Question

18. If there are (2n + 1) terms in an arithmetic series, then prove that the ratio of the sum of odd terms to the sum of even terms is (n+1):n.

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

Answer 2

Solution :

Let a and d be the first term and

common difference respectively

of the given Arithmetic Progression.

Let am denote the m th terms of

the A.P .

am = a + ( m - 1 )d

S1 = Sum of odd terms

= a1 + a3 + ...+ a2n+1

= ( n+1)/2[ a1 + a2n+1]

= ( n+1)/2 [ a+a+(2n+1-1)d ]

**********************************

a2n+1 = a + (2n-1-1)d

********************†*************

S1 = ( n + 1 )( a + nd ) ----( 1 )

And

S2 = Sum of even terms

= a2 + a4 + ....+ a2n

= n/2 [ a2 + a2n ]

= n/2 { (a+d)+[a+(2n-1)d] }

[ Since , a2n = a + (2n-1)d ]

S2 = n(a + nd) -----( 2 )

Therefore ,

S1 : S2 = (n+1)(a+nd) : n(a+nd)

= ( n + 1 ) : n

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