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Question
If there are (2n + 1) terms in an A.P., then prove that the ratio of the sum of odd terms and the sum of even terms is (n + 1) : n
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Solution
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.
Then S1 = a1 + a3 + a5 + ... + a2n+1
S1 = `(n + 1)/2 (a_1 + a_(2n + 1))`
S1 = `(n + 1)/2 [a + a + (2n + 1 - 1)d]`
= (n + 1) (a + nd)
Similarly, if S2 denotes the sum of even terms, then
S2 = `n/2 [2a + 2nd]` = n(a + nd)
Hence, `"S"_1/"S"_2 = ((n + 1)(a + nd))/(n(a + nd))`
= `(n + 1)/n`
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