Question

The ratio of the sum of my and n term of a ap is m2 and n2. Show that the ratio of the Mth and Nth is (2m-1):(2n-1)

Answer 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]

→ {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)

Answer 2

||✪✪ QUESTION ✪✪||

The ratio of the sum of m and n term of a ap is m² and n². Show that the ratio of the Mth and Nth is (2m-1):(2n-1). ?

|| ★★ FORMULA 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)

|| ✰✰ ANSWER ✰✰ ||

Given That, The ratio of the sum of m and n term of a ap is m² and n²,,

So,

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

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

Now, Denominator 2 will be cancel From Denominator 2 in LHS, and m² will be cancel with m , and n² will be cancel with n,,

we left now,

→ 2a + (m-1)d / 2a + (n-1)d = m /n

Cross - Multiply,

→ 2an + (m-1)dn = 2am + (n-1)dm

→ 2an - 2am = d[ (n-1)m - (m-1)n ]

→ 2a(n - m) = d [ (mn - m) - (mn - n) ]

→ 2a(n - m) = d [(mn - m - mn + n) ]

→ 2a(n - m) = d ( n - m)

→ 2a = d

____________________________

Now, By above Told Formula :-

→ mth term of AP = a + (m-1)d

→ nth term of AP = a + (n-1)d.

Putting value of d = 2a in both ,, and than Checking their Ratio

→ [ a + (m-1)2a ] : [ a + (n-1)2a ]

Taking a common From both,

→ a [ 1 + (m-1)2 ] : a [ 1 + (n - 1)2 ]

a will be cancel From both sides ,

→ ( 1 + 2m - 2) : ( 1 + 2n - 2)

→ (2m - 1) : (2n - 1) .

✪✪ Hence Proved ✪✪

So, we can say that, the ratio of the Mth and Nth is (2m-1):(2n-1) .