Question

If s1 be the sum of (2n+1) terms of an a.p. and s2 be the sum of odd terms, then prove that s1:s2=(2n+1):(n+1)

Answer 1

Answer:

s1: s2 :: (2n + 1) :  (n+1)

Step-by-step explanation:

Let say an AP is

a  , a+d , a + 2d  , ....................................a + 2nd

s1 = ((2n + 1)/2)(a + a + (2n + 1 - 1)d)

s1 = ((2n + 1)/2)(2a + 2nd)

s1 = (2n + 1)(a + nd)

AP of odd terms will be

a  a + 2d  ,  a + 4d  ,.....................................a + 2nd

n +1  terms

d = 2d

s2 = ((n+1)/2)(a + a + (n+1-1)2d)

s2 = (n+1)/2(2a + 2nd)

s2 = (n+1)(a + nd)

s1: s2 ::  (2n + 1)(a + nd) : (n+1)(a + nd)

=> s1: s2 :: (2n + 1) :  (n+1)

QED

Proved

Answer 2

Answer:Let the given A.P. has (2n + 1) terms

So, the sum of (2n + 1) terms is i.e. S1,

2n+1/2{2a + (2n+1 - 1)d

= 2n+1/2 × 2{a + nd}

= 2n+1{a + nd} ----equ. 1

And the sum of odd terms i.e. S2,

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

(since, no. of odd terms is (n+1) and common difference is 2d)

= n+1/2 × 2{a + nd}

= n+1{a + nd} ----equ. 2

Now, Dividing equ. 1 by 2, we get

S1/S2 = 2n+1/n+1. Hence Proved

Step-by-step explanation: