Question

if the ratio of the sum of m terms and n terms of an ap b m square is 2 m square prove that the ratio of its 10th term is 2 m - 1 is to 2 and -1​

Answer 1

Correct 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

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 cancel throughout

⇒ (1 + 2m - 2)/(1 + 2n - 2)

⇒ (2m - 1)/(2n - 1)

Hence, proved

Answer 2

||✪✪ GIVEN ✪✪||

• Ratio of sum of m and n terms of an AP = m² : n² .
• Show That ratio of mth and nth term = (2m - 1) : (2n -1) .

|| ★★ CONCEPT USED ★★ ||

• A sequence is said to be in AP (Arithmetic Progression), if the difference between its consecutive terms are equal.

• The nth term of an AP is given as :-

→ T(n) = a + (n-1)•d , where a is the first term and d is the common difference.

• The common difference of an AP is given as :-

→ d = T(n) - T(n-1)

• If the number of terms in an AP is n ( where n is odd ) ,then there will be a single middle term.

Also, [(n+1)/2]th term will be its middle term.

• If the number of terms in an AP is n ( where n is even ) ,then there will be two middle terms.

Also, (n/2)th and (n/2 + 1)th terms will be its middle terms.

• The sum up to nth terms of an AP is given as :-

→ S(n) = (n/2)•[2a + (n-1)•d] where a is the first term and d is the common difference.

• The nth term of an AP is also given as :-

T(n) = S(n) - S(n-1)

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|| ✰✰ ANSWER ✰✰ ||

Let The AP Series First Term is a, Common Difference is d.

As Told Above Formula :-

➼ sum of m terms of AP = (m/2)[2a + (m - 1)d ]

And,

➼ sum of n terms of AP = (n/2)[2a + (n -1)d ]

Now, Given That, Their Ratio is m² : n² .

So,

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

Or,

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

m & n cancel from LHS and RHS, and Denominator 2 , will also cancel Each other ,

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

Cross - Multiply Now, we get,

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

➪ n * [2a + md - d] = m * [2a + nd - d]

➪ 2an + mdn - dn = 2am + mdn - dm

cancelling mdn from both sides ,

➪ 2an - 2am = dn - dm

➪ 2a(n - m) = d(n - m)

Cancelling (n - m) from both sides ,

➪ 2a = d --------------- Equation ❶

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Now, we Have To Find Ratio of m(th) term and n(th) Term.

☞ m(th) Term of This AP = a + (m - 1)d .

☞ n(th) Term of This AP = a + (n - 1)d .

Required Ratio :-

☛ a + (m - 1)d : a + (n - 1)d

Putting Value of d From Equation ❶ Now,

☛ a + (m - 1)(2a) : a + (n - 1)(2a)

☛ a + 2am - 2a : a + 2an - 2a

☛ 2am - a : 2an - a

☛ a(2m - 1) : a(2n - 1)

Cancelling a now,

☛ (2m - 1) : (2n - 1). ✪✪ Hence Proved ✪✪

ஃ we can say That, Ratio of m(th) term and n(th) term of Given AP will be (2m - 1) : (2n - 1)..

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