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Question: In an arithmetic progression if \(m\left( {{a_m}} \right) = n\left( {{a_n}} \right)\). Find the valu...

In an arithmetic progression if m(am)=n(an)m\left( {{a_m}} \right) = n\left( {{a_n}} \right). Find the value of (am+n)th{\left( {{a_{m + n}}} \right)^{th}} term.

Explanation

Solution

Here, the mth{m^{th}} and the nth{n^{th}} term of an arithmetic progression satisfy the relation m(am)=n(an)m\left( {{a_m}} \right) = n\left( {{a_n}} \right). we have to calculate the (m+n)th{\left( {m + n} \right)^{th}} term of the same arithmetic progression. The nth{n^{th}} term of an arithmetic progression is given by a formula an=a+(n−1)d{a_n} = a + \left( {n - 1} \right)d, where aa is the first term, dd is the common difference and nn is the number of term of the given A.PA.P.

Complete step-by-step solution:
The given relation between the mth{m^{th}} and the nth{n^{th}} terms is m(am)=n(an)m\left( {{a_m}} \right) = n\left( {{a_n}} \right).
We know that the nth{n^{th}} term of an arithmetic progression is given by an=a+(n−1)d{a_n} = a + \left( {n - 1} \right)d.
Suppose the first term of an arithmetic progression is aa and the common difference is dd, we have to find the value of the nth{n^{th}} and the mth{m^{th}} terms. So, by applying above given formula we get,
am=a+(m−1)d{a_m} = a + \left( {m - 1} \right)d.
an=a+(n−1)d{a_n} = a + \left( {n - 1} \right)d.
Now, putting these values in the given relation, we get,
\Rightarrow m\left\\{ {a + \left( {m - 1} \right)d} \right\\} = n\left\\{ {a + \left( {n - 1} \right)d} \right\\} \\\ \Rightarrow ma + m\left( {m - 1} \right)d = na + n\left( {n - 1} \right)d \\\ \Rightarrow ma - na = \left( {{n^2} - n} \right)d - \left( {{m^2} - m} \right)d
Taking aa as a common from the terms on the RHS and dd as a common from the terms on the left hand side. We get,
⇒a(m−n)=d(n2−n−m2+m) ⇒−a(n−m)=d(n2−m2−n+m) \Rightarrow a\left( {m - n} \right) = d\left( {{n^2} - n - {m^2} + m} \right) \\\ \Rightarrow - a\left( {n - m} \right) = d\left( {{n^2} - {m^2} - n + m} \right)
Now, applying the formula a2−b2=(a−b)(a+b){a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right). We get,
\Rightarrow - \left( {n - m} \right) = d\left\\{ {\left( {n - m} \right)\left( {n + m} \right) - \left( {n - m} \right)} \right\\}
Taking (n−m)\left( {n - m} \right)from the terms on the RHS. We get,
⇒−a(n−m)=d(n−m)(m+n−1) ⇒−a=d(m+n−1) \Rightarrow - a\left( {n - m} \right) = d\left( {n - m} \right)\left( {m + n - 1} \right) \\\ \Rightarrow - a = d\left( {m + n - 1} \right)
⇒∴a=−d(m+n−1)\Rightarrow \therefore a = - d\left( {m + n - 1} \right) - - - - - - - - - - - - -(1)
Now, we have to calculate the (m+n)th{\left( {m + n} \right)^{th}} term of the arithmetic progression. By applying above given formula we get,
am+n=a+(m+n−1)d{a_{m + n}} = a + \left( {m + n - 1} \right)d.
From the equation (1) we get the value of a=−d(m+n−1)a = - d\left( {m + n - 1} \right). Putting the value of aa, we get,
⇒am+n=−d(m+n−1)+(m+n−1)d ∴am+n=0 \Rightarrow {a_{m + n}} = - d\left( {m + n - 1} \right) + \left( {m + n - 1} \right)d \\\ \therefore {a_{m + n}} = 0

Thus, we get the value of the (m+n)th{\left( {m + n} \right)^{th}} term of AP is zero.

Note: The summation of the nn terms of an arithmetic progression is given by the formula {S_n} = \dfrac{n}{2}\left\\{ {2a + \left( {n - 1} \right)d} \right\\} where aa and dd are the first term and the common difference of an AP.