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The ratio of sum of m and n terms of an AP is m sq. :n sq. show that ratio of mth and nth term is m/n=2m-1/2n-1. solve the question?
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The ratio of sum of m and n terms of an AP is m sq. :n sq. show that r...
Sum of m terms of an A.P. = m/2 [2a + (m -1)d]
Sum of n terms of an A.P. = n/2 [2a + (n -1)d]
m/2 [2a + (m -1)d] / n/2 [2a + (n -1)d] = m:n
⇒ [2a + md - d] / [2a + nd - d] = m/n
⇒ 2an + mnd - nd = 2am + mnd - md
⇒ 2an - 2am = nd - md
⇒ 2a (n -m) = d(n - m)
⇒ 2a = d
Ratio of m th term to nth term:
[a + (m - 1)d] / [a + (n - 1)d]
= [a + (m - 1)2a] / [a + (n - 1)2a]
= a [1 + 2m - 2] / a[1 + 2n -2]
= (2m - 1) / (2n -1)
So, the ratio of mth term and the nth term of the arithmetic series is (2m - 1):(2n -1).
Community Answer
The ratio of sum of m and n terms of an AP is m sq. :n sq. show that r...
Explanation:
Given:
Sum of m and n terms of an AP= m sq. :n sq.
To prove:
Ratio of mth and nth term is m/n=2m-1/2n-1
Formula:
Sum of first n terms of an AP= n/2[2a+(n-1)d]
Sum of first m terms of an AP= m/2[2a+(m-1)d]
mth term of an AP= a+(m-1)d
nth term of an AP= a+(n-1)d
Proof:
Given, Sum of m and n terms of an AP= m sq. :n sq.
So, m/2[2a+(m-1)d] : n/2[2a+(n-1)d] = m sq. :n sq.
On simplifying, we get,
(2a+(m-1)d)/(2a+(n-1)d) = m/n
On cross multiplying, we get,
mn(2a+(m-1)d)=nm(2a+(n-1)d)
On solving, we get,
md+nd-m-n=0
d(m+n)=m-n
d=(m-n)/(m+n)
Now, we can find the mth and nth term of the AP.
mth term of an AP= a+(m-1)d
Substituting the value of d, we get,
mth term of an AP = a + (m-1)(m-n)/(m+n)
nth term of an AP= a+(n-1)d
Substituting the value of d, we get,
nth term of an AP = a + (n-1)(m-n)/(m+n)
Now, we can find the ratio of mth and nth term.
Ratio of mth and nth term = (a + (m-1)(m-n)/(m+n))/(a + (n-1)(m-n)/(m+n))
On simplifying, we get,
Ratio of mth and nth term = m/n=2m-1/2n-1
Conclusion:
Hence, the ratio of mth and nth term is m/n=2m-1/2n-1
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The ratio of sum of m and n terms of an AP is m sq. :n sq. show that ratio of mth and nth term is m/n=2m-1/2n-1. solve the question?
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