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:-

Given:-

• Ratio of sum of n terms and n terms of an AP = m² : n².

To Prove:-

• d = 2a.

Formula Used:-

$\tt \blue{\leadsto{ \boxed{ \red{ \bf S_x = \frac{a}{2} \bigg(2a + (x - 1)d \bigg).}}}}$

Where,

• $\tt S_x =$ Sum of x terms.

• a = First term.

• x = Number of terms.

• d = Common Difference

So Here,

• $\tt S_x = S_m\:\:And\:\:S_n.$

• a = a (lets assume as per question).

• x = m and n.

• d = d (lets assume as per question).

Now,

By putting the above values in formula we get,

$\tt\longrightarrow \boxed{ \bf S_m = \frac{m}{2} \bigg(2a + (m - 1)d \bigg)}...(1)$

And,

$\tt\longrightarrow \boxed{ \bf S_n = \frac{n}{2} \bigg(2a + (n - 1)d \bigg)}...(2)$

Therefore ATQ:-

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

So from (1) and (2) we get,

$\tt\mapsto \dfrac{ \dfrac{m}{ \cancel2} \bigg(2a + (m - 1)d \bigg) }{ \dfrac{n}{ \cancel2} \bigg(2a + (n - 1)d \bigg) } = \dfrac{ {m}^{2} }{ {n}^{2} }$

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

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

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

$\tt\mapsto n \bigg(2a + (m - 1)d \bigg) = m \bigg(2a + (n - 1)d \bigg).$

$\tt\mapsto n(2a + nd - d) = m(2a + md - d).$

$\tt\mapsto2an \: \: \cancel{ + mnd} \: \: - nd = 2am \: \: \cancel{ + mnd} \: \: - md.$

$\tt\mapsto2an - nd = 2am - md.$

$\tt\mapsto md - nd = 2am - 2an.$

$\tt\mapsto d \: \: \cancel{(m - n)} = 2a \: \: \cancel{(m - n)}.$

$\tt \large\red{\mapsto{ \boxed{ \blue{ \bf d = 2a.}}}}$

...Hence Proved...