Question

If the ratio of the sum of m term and n term of an ap is m²:n², then prove that d=2a

Answer 1

Answer:

$\Large{\underline{\underline{\bf{Given:}}}}$

• Ratio of sum of m terms and n terms of an ap is m² : n²

$\Large{\underline{\underline{\bf{To\:prove:}}}}$

• d = 2a

$\Large{\underline{\underline{\bf{Proof:}}}}$

→ Sum of n terms of an AP is given by the formula

$S_n=\dfrac{n}{2}( 2a+(n-1)d)$ ----equation 1

→ By given,

$\dfrac{S_m}{S_n} =\dfrac{m^{2} }{n^{2} }$

→ Substituting equation 1

$\dfrac{\dfrac{m}{2}(2a+(m-1)d) }{\dfrac{n}{2}(2a+(n-1)d) }=\dfrac{m^{2} }{n^{2} }$

→ Cancelling 2 on both numerator and denominator

$\dfrac{m(2a+(m-1)d)}{n(2a+(n-1)d)} =\dfrac{m^{2} }{n^{2} }$

→ Taking m/n to RHS

$\dfrac{2a+(m-1)d}{2a+(n-1)d} =\dfrac{m^{2} }{n^{2} } \times \dfrac{n}{m}$

$\dfrac{2a+(m-1)d}{2a+(n-1)d} =\dfrac{m}{n}$

→ Cross multiplying

$n(2a+(m-1)d)=m(2a+(n-1)d)$

$2an+(m-1)dn = 2am+(n-1)dm$

$(m-1)dn-(n-1)dm=2am-2an$

$dmn-dn-dmn+dm=2a(m-n)$

→ Cancelling dmn

$dm-dn=2a(m-n)$

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

→ Cancelling m - n on both sides

d = 2a

Hence proved.

$\Large{\underline{\underline{\bf{Notes:}}}}$

→ Sum of n terms is given by the equations

$S_n=\dfrac{n}{2} (a_1+a_n)$

$S_n=\dfrac{n}{2}( 2a+(n-1)d)$

Answer 2

EXPLANATION.

• GIVEN

ratio of sum of m term and n term of

an Ap = m²:n²

Prove that d = 2a.

According to the question,

Formula of sum of Nth term of an Ap.

$\implies{\boxed{ \green{\bold{s_{n} \: = \frac{n}{2} (2a \: + (n \: - 1)d) }}}}$

$\bold{\implies{ s_{m} \: = {m}^{2} } }$

$\bold{\implies{ s_{n} \: = {n}^{2} }}$

Therefore,

$\bold{\implies{ \frac{ \frac{m}{2} (2a \: + (m \: - 1)d)}{ \frac{n}{2}(2a \: + (n \: - 1)d) } } = \frac{ {m}^{2} }{ {n}^{2} } }$

$\bold{\implies{ \frac{2a \: + \: ( \: m \: - \: 1)d}{2a \: + \: (n \: - 1)d} } = \frac{m}{n}}$

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

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

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

=> 2an - 2am = nd - md

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

=> 2a = d

=> HENCE PROVED.