In the page arithmetic series worksheet solution7 you are going to see solution of each questions from the arithmetic series worksheet.

(18) If there are (2n+1) terms in an arithmetic series,then prove that the ratio of the sum of odd terms to the sum of even terms is (n+1) : n

**Solution:**

Let T and S are sum of odd terms and even numbers respectively.

T = t₁ + t₃ + t₅ + ........... + t(2n+1)

there are (n+1) terms. To find its sum we have to use Sn formula but here Sn is Tn

Tn = (n/2) [a + L]

= [(n+1)/2] [ t₁ + t(2n+1)]

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

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

= [(n+1)/2] [ 2a + (2n)d]

= [(n+1)] [ a + n d ]

S = t₂ + t₄ + t₆ + ........... + t2n

Sn = (n/2) [a + L]

= (n/2)[ t₂ + t 2n]

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

= (n/2)[a+d + a + 2nd-d]

= (n/2)[2a+2nd]

= n [a + n d]

= sum of odd terms : sum of even terms

= [(n+1)] [ a + n d ] : n[a + n d]

= [(n+1)] [ a + n d ]/n[a + n d]

= [(n+1)]/n

= (n+1):n

(19) The ratio of the sums of first m and first n terms of an arithmetic series is m²:n² show that the ratio of the m th and nth terms is (2m-1) :(2n-1)

**Solution:**

Sum of m terms:Sum of n terms = m²:n²

(m/2)[2a+(m-1)d]/(n/2)[2a+(n-1)d] = m²/n²

(m/2) x (2/n) [2a+(m-1)d]/[2a+(n-1)d] = m²/n²

(m/n) [2a+(m-1)d]/[2a+(n-1)d] = m²/n²

[2a+(m-1)d]/[2a+(n-1)d] = m/n

n[2a+(m-1)d] = m[2a+(n-1)d]

2na+ n (m-1)d = 2ma + m (n-1) d

2na+ n m d - n d = 2ma + m n d - m d

2na - 2ma = n d - m d

2 a (n-m) = d (n-m)

2 a = d

Now we have to verify that the ratio of the m th and nth terms is (2m-1) :(2n-1)

tm:tn = [a + (m-1) d]/[ a + (n-1) d]

= [a + (m-1) (2a)]/[ a + (n-1) (2a)]

= [a + 2ma-2a]/[a + 2na-2a]

= [-a + 2ma]/[-a + 2na]

= a [2m - 1]/a[2n - 1]

= [2m - 1]/[2n - 1]

= [2m - 1]:[2n - 1]

- Arithmetic series worksheet
- Special series
- Sequence
- Arithmetic progression
- Arithmetic series
- Geometric progression
- Geometric series

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arithmetic series worksheet solution7