Question

The sum of m terms and n terms of an ap is m2n2 then show that it's mth term and nth term are in the ratio (2m-1) : (2n-1)​

Answer 1

Answer:

Step-by-step explanation:

Sum of m terms of an A.P. = m²

⇒ m² = $\frac{m}{2}[2a+(n-1)d]$

Sum of n terms of an A.P.

⇒ n² = $\frac{n}{2}[2a+(n-1)d]$

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

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

⇒ n( [2a + md - d] ) : m( [2a + nd - d] ) = m : n

⇒ 2an + $mnd$ - nd = 2am + $mnd$ - md

Cancel $mnd$ on both sides

⇒ 2an - 2am = nd - md

⇒  2a (n -m) = d(n - m)

⇒  2a = d

m th term = a+(m-1)d

n th term = a+(n-1)d

m : n = (2m - 1) : (2n -1)

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

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

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

⇒ (2m - 1) : (2n -1) = (2m - 1) : (2n -1)

LHS = RHS