Ratio of mth and nth terms of an A. P. with given ratio of sums
Given that the ratio to sum of first m and n terms of an A.P. with first term ‘a’ and commond difference ‘d’ is m^2:n^2. The task is to find the ratio of mth and nth term of this A.P.
Input: m = 3, n = 2 Output: 1.6667 Input: m = 5, n = 3 Output: 1.8
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Let the Sum of first m and n terms be denoted by Sm and Sn respectively.
Also, let the mth and nth term be denoted by tm and tn respectively.
Sm = (m * [ 2*a + (m-1)*d ])/2
Sn = (n * [ 2*a + (n-1)*d ])/2
Given: Sm / Sn = m^2 / n^2
Hence, ((m * [ 2*a + (m-1)*d ])/2) / ((n * [ 2*a + (n-1)*d ])/2) = m^2 / n^2
=> (2*a + (m-1)*d) / (2*a + (n-1)*d) = m / n
on cross multiplying and solving, we get
d = 2 * a
Hence, the mth and nth terms can be written as:
mth term = tm = a +(m-1)*d = a + (m-1)*(2*a)
nth term = tn = a +(n-1)*d = a + (n-1)*(2*a)
Hence the ratio will be:
tm / tn = (a + (m-1)*(2*a)) / (a + (n-1)*(2*a))
tm / tn = (2*m – 1) / (2*n – 1)
Below is the required implementation: