Question

the ratio of the sum of the first M terms to n terms is m square: 2 m square then show that the ratio of the mth term to the nth term is (2m-1):(2n-1)​

Answer 1

Answer:

Sum of m terms of an A.P. = m/2 [2a + (m -1)d]

Sum of n terms of an A.P. = n/2 [2a + (n -1)d]

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

⇒ [2a + md - d] / [2a + nd - d] = m/n

⇒ 2an + mnd - nd = 2am + mnd - md

⇒ 2an - 2am = nd - md

⇒ 2a (n -m) = d(n - m)

⇒ 2a = d

Ratio of m th term to n th term:

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

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

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

= (2m - 1) / (2n -1)

So, the ratio of m th term and the n th term of the arithmetic series is (2m - 1) : (2n -1).

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Answer 2

Question

The ratio of the sums of first m and n terms of an ap is m²:n². Show that the ratio of the m and n term is (2m-1) :(2n-1)

To show

(2m-1):(2n-1)

Solution

Used formulas:

• Sum of m terms of an A.P. = m/2 [2a + (m -1)d]
• Sum of n terms of an A.P. = n/2 [2a + (n -1)d]

According to question,

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

=> 2[2a + md - d]/2[2a + nd - d] = m/n

=> (2a + md - d)/(2a + nd - d) =m/n

=> n(2a + md - d) = m(2a + nd - d)

=> 2an + mnd - nd = 2am + mnd - md

=> 2an - 2am + mnd - mnd = - md +nd

=> 2an - 2am = - md + nd

Take 2a common on LHS and d on RHS

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

=> d = 2a

Also, ratio of m th term to n th term,

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

Substitute value of d from above

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

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

a throughout cancel

=> (1 + 2m - 2)/(1 + 2n - 2)

=> (2m - 1)/(2n - 1)

Hence, proved