Question

If there are (2n + 1) terms in an 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

We have to prove the ratio of sum of odd terms to sum of even terms of an arithmetic series to be

n+1/n

.

Since there are 2n+1 terms there will be n+1 terms.

Let us consider the odd terms and even terms to be two different series.

These series will have common difference 2d, where d is the common difference of original series.

Let a be the first term.

Sum of odd term series = n+1/2 (2a+n×2d) ....(i)

Sum of even term series = n/2(2(a+d)+(n−1)×2d)=

dn/2(2a+n×2d) ....(ii)

The ratio

n+1/2(2a+n×2d) n+1

_____________ = ____ = n+1 : n

n/2(2a+n×2d) n

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Answer 2

$\green{\large\underline{\sf{Solution-}}}$

Given that there are 2n + 1 terms in an AP series.

Let assume that the terms be

$\rm :\longmapsto\:a_1,a_2,a_3, - - - - ,a_{2n}, \: a_{2n + 1}$

with first term a and common difference d respectively.

So, terms at odd places be

$\rm :\longmapsto\:a_1,a_3,a_5, - - - - , \: a_{2n + 1}$

So, sum of terms at odd places be

$\rm :\longmapsto\:S_{odd} \: = \: a_1 + a_3 + a_5 + - - - - + \: a_{2n + 1}$

$\rm :\longmapsto\:S_{odd} \: = \: a + (a + 2d) + (a + 4d)+ - - (n + 1) \: terms$

Here,

First term of AP series = a

Common difference of an AP = 2d

Number of terms = n + 1

So,

$\rm :\longmapsto\:S_{odd} = \dfrac{n + 1}{2} \bigg(2a + (n + 1 - 1)2d \bigg)$

$\rm :\longmapsto\:S_{odd} = \dfrac{n + 1}{2} \bigg(2a + 2nd \bigg)$

$\rm :\longmapsto\:\boxed{ \tt{ \: S_{odd} = (n + 1) \bigg(a + nd \bigg)}} - - - (1)$

Now, terms at even places are

$\rm :\longmapsto\:a_2,a_4,a_6, - - - - , \: a_{2n}$

So, Sum of terms at even places is

$\rm :\longmapsto\:S_{even} \: = \: a_2 + a_4 + a_6 + - - - - + \: a_{2n}$

$\rm :\longmapsto\:S_{even} \: = \: (a + d) + (a + 3d) + (a + 5d)+ - - n\: terms$

Here,

First term of AP series = a + d

Common difference of an AP = 2d

Number of terms = n

So,

$\rm :\longmapsto\:S_{even} = \dfrac{n}{2} \bigg(2(a + d) + (n - 1)2d\bigg)$

$\rm :\longmapsto\:S_{even} = \dfrac{n}{2} \bigg(2a + 2d + 2nd - 2d\bigg)$

$\rm :\longmapsto\:S_{even} = \dfrac{n}{2} \bigg(2a + 2nd \bigg)$

$\rm :\longmapsto\:\boxed{ \tt{ \: S_{even} =n(a + nd) \: }} - - - - (2)$

Hence,

$\red{\rm :\longmapsto\:S_{odd} \: : \: S_{even}}$

$\rm \:  =  \: (n + 1)(a + nd) \: : \: n(a + nd)$

$\rm \:  =  \: n + 1 \: : \: n$

Therefore,

$\red{\rm :\longmapsto\:S_{odd} \: : \: S_{even} \: = \: n \: + \: 1 \: : \: n}$

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Additional Information :-

↝ Sum of n  terms of an arithmetic sequence is,

$\begin{gathered}\red\bigstar\:\:{\underline{\orange{\boxed{\bf{\green{S_n\:=\dfrac{n}{2} \bigg(2 \:a\:+\:(n\:-\:1)\:d \bigg)}}}}}} \\ \end{gathered}$

Wʜᴇʀᴇ,

• Sₙ is the sum of n terms of AP.

• a is the first term of the sequence.

• n is the no. of terms.

• d is the common difference.

↝ nᵗʰ term of an arithmetic sequence is,

$\begin{gathered}\red\bigstar\:\:{\underline{\orange{\boxed{\bf{\green{a_n\:=\:a\:+\:(n\:-\:1)\:d}}}}}} \\ \end{gathered}$

Wʜᴇʀᴇ,

• aₙ is the nᵗʰ term.

• a is the first term of the sequence.

• n is the no. of terms.

• d is the common difference.