if the ratio of the sum of first m and n terms of AP is m2:n2 show that the ratio of its mth and nth term is (2m-1):(2n-1)
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Answer:
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Answer:
Step-by-step explanation:
Solution :-
Let the a be first term and d be the common difference and sum of first m term and n term be S(m) and S(n) respectively.
Therefore,
S(m)/S(n) = m²/n²
⇒ m/2[2a + (m - 1)d]/n/2[2a + (n - 1)d] = m²/n²
⇒ 2a + (m - 1)d/2a + (n - 1)d = m²/n² × n/m
⇒ 2a + (m - 1)d/2a + (n - 1)d = m/n
⇒ m[2a + (n - 1)d] = n[2a + (m - 1)d]
⇒ d = 2a
Now,
⇒ a(m)/a(n) = a + (m - 1)d/a + (n - 1)d
⇒ a(m)/a(n) = a + (m - 1) (2d)/a + (n - 1) (2a)
⇒ a(m)/a(n) = 2a + 2ma - 2a/2a + 2na - 2a
⇒ a(m)/a(n) = 2ma - a/2na - a
⇒ a(m)/a(n) = a(2m - 1)/a(2n - 1)
⇒ a(m)/a(n) = 2m - 1 : 2n - 1
Hence, Proved.
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