Question

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

Answer 1

let the first term of AP be a and the common difference be d.

the number of terms in AP is 2n+1.

the terms are a,a+d,a+2d,..........,a+(2n-1)d, a+2nd.

last term will be odd term. the number of odd terms are (2n+1+1)/2=(2n+2)/2=n+1 terms

odd terms are a, a+2d,.........,a+2nd

the sum of odd terms is
So= (n+1)/2(a+a+2nd)
So= (n+1)/2(2a+2nd)
So=(n+1)(a+nd) Eqn1.

the number of even terms is 2n+1-(n+1)= n terms

even terms are a+d,a+3d,.................a+(2n-1)d

the sum of even terms is
Se= n/2 [a+d+a+(2n-1)d]
Se= n/2 [2a+2nd]
Se= n (a+nd) Eqn2.


from (1) and (2) the ratio of sum of odd terms and the sum of even terms is
S(o)/S(e)= (n+1)(a+nd)/ (n)(a+nd)
Therefore, (n+1)/n
Here o= odd and e= even

, which is required result.

Answer 2

Answer:

Step-by-step explanation:

See attachment for answer