Question

If S₁=2+4+...+2n and S₂ = 1+3+...+(2n–1), then S₁ : S₂ = .....,select a proper option (a), (b), (c) or (d) from given options so that the statement becomes correct.(All the problems refer to A.P.)
(a)n+1/n
(b) n/n+1
(c) n²
(d) (n+1)

Answer 1

S₁ = 2 + 4 + ... + 2n
first term, a = 2 , common difference, d = 2 and last term, Tx = 2n.
use formula, Tx = a + (x -1)d
2n = 2 + (x - 1)2
2n = 2 + 2x - 2
x = n , hence number of terms = n
now, use formula , $S_x=\frac{x}{2}[a+T_x]$
here, x = n , Tx = 2n , a = 2
so, S₁ = n/2 [ 2 + 2n ]
= n(n + 1) .......(i)

S₂ = 1 + 3 + ... + (2n – 1)
first term, a' = 1 , common difference , d' = 2
and last term , T'x = (2n - 1)
use formula, T'x = a' + (x - 1)d'
2n - 1 = 1 + (x - 1)2
2n - 1 = 1 + 2x - 2
x = n , hence number of terms = n
now use formula, $S_x=\frac{x}{2}[a'+T'_x]$
here, x = n, a' = 1 , T'x = (2n - 1)
S₂ = n/2 [1 + 2n - 1]
= n² ........(ii)

hence, $\frac{S_1}{S_2}=\frac{n(n+1)}{n^2}$
$\frac{S_1}{S_2}=\frac{n+1}{n}$

hence, option (a) is correct.

Answer 2

$S_{1} = 2 + 4 + ... + 2n \\ S_{2} = 1 + 3 + ... + (2n - 1) \\ \\ S_{1} = 2(1 + 2 + ... + n) \\ S_{1} = n(n + 1) \\ \\ S_{2} = {n}^{2} \: ( \: Using \: identity \: ) \\ \\ \frac{S_{1}}{S_{2}} = \frac{n(n + 1)}{ {n}^{2} } \\ \\ \boxed{ \large \: \frac{S_{1}}{S_{2}} = \frac{n + 1}{n} }$