If S1 be the sum of (2n + 1) terms of an A.P. and S2 sum of its odd te

1.	If S1 be the sum of (2n + 1) terms of an A.P. and S2 sum of its odd terms, then prove that: S1: S2 = (2n + 1) : (n + 1).
Answer» To prove : S₁: S₂ = (2n + 1) : (n + 1) We know that the sum of AP is given by the formula : s = \(\frac{n}{2}\)(2a + (n-1)d) Substituting the values in the above equation, s₁ = \(\frac{2n+1}{2}\)(2a+2nd) For the sum of odd terms, it is given by, s₂ = a₁ + a₃ + a₅ + …. a_{2n + 1} s₂ = a + a + 2d + a + 4d + … + a + 2nd s₂ = (n + 1)a + n(n + 1)d s₂ = (n + 1) (a + nd) Hence, s₁: s₂= \(\frac{2n+1}{n+1}\)

If S1 be the sum of (2n + 1) terms of an A.P. and S2 sum of its odd terms, then prove that: S1: S2 = (2n + 1) : (n + 1).

Answer»

To prove : S₁: S₂ = (2n + 1) : (n + 1)

We know that the sum of AP is given by the formula :

s = \(\frac{n}{2}\)(2a + (n-1)d)

Substituting the values in the above equation,

s₁ = \(\frac{2n+1}{2}\)(2a+2nd)

For the sum of odd terms, it is given by,

s₂ = a₁ + a₃ + a₅ + …. a_{2n + 1}

s₂ = a + a + 2d + a + 4d + … + a + 2nd

s₂ = (n + 1)a + n(n + 1)d

s₂ = (n + 1) (a + nd)

Hence,

s₁: s₂= \(\frac{2n+1}{n+1}\)