Question

If there are (2n+1) terms in A.P., then prove that the ratio of the sum of odd terms and the sum of even terms is (n+1):n.

Answer 1

Step-by-step explanation:

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

Answer 2

SOLUTION

Let ap be the pth term of the given A.P.Then,

ap= a+(p-1)d

Now,

s1= sum of odd terms

s1= a1+a3+a5+.....+a2n+1

s1= n+1/2{a1+ a2n+1}

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

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

and

s2= sum of even terms

s2= a2+ a4+ a6+....+a2n

s2= n/2[a2+a2n]

s2= n/2[(a+d)+{a+(2n-1)d}]

s2= n(a+nd)

Hence,

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

hope it helps ☺️