Question

$hello \: mates \:$

$the \: ratio \: of \: the \: sum \: of \: m \\ and \: n \: terms \: of \: an \: a.p. \: is \\ m {}^{2} ratio \: n {}^{2}. \\ show \: that \: the \: ratio \: of \: m {}^{th}$
$and \: n {}^{th} \: term \: is \: \\ (2m - 1)ratio(2n - 1)$


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Answer 1

let a be first term and d be common difference

Sm= m/2. ( 2a + ( m-1)d)

Sn = n/2. (2a + ( n-1) d)


Sm/Sn = m/n . ( 2a + ( m-1)d)/ ( 2a +( n-1)d)

m^2/n^2 = m/n. ( 2a + ( m-1) d)/ ( 2a +( n-1)d)

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


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

Compare

d = 1

2a - d = 0

2a = 1

a= 1/2


am = a + ( m-1)d

= 1/2 + ( m-1)

= m -1/2

an = n -1/2

am /an = m - 1/2)/ n -1/2

= 2 m -1)/ 2n -1