Let the Arithmetic Progression (AP) have the first term \(a\) and the common difference \(d\).
The formula for the \(n^{th}\) term of an AP is given by: \(a_n = a + (n-1)d\)
We are given that the \(p^{th}\) term of the AP is \(k\). Using the formula:
\(a_p = a + (p-1)d = k \quad (1)\)We need to find the sum of the \(p^{th}\) term, \((p+q)^{th}\) term, and \((p-q)^{th}\) term. Let's express these terms:
The sum \(S\) is:
\(S = a_p + a_{p+q} + a_{p-q}\) \(S = [a + (p-1)d] + [a + (p+q-1)d] + [a + (p-q-1)d]\)Combine like terms:
\(S = (a + a + a) + [(p-1)d + (p+q-1)d + (p-q-1)d]\) \(S = 3a + d[(p-1) + (p+q-1) + (p-q-1)]\)Simplify the expression inside the brackets:
\((p-1) + (p+q-1) + (p-q-1) = p - 1 + p + q - 1 + p - q - 1 = 3p - 3\)Substitute this back into the sum equation:
\(S = 3a + d(3p - 3)\) \(S = 3a + 3(p-1)d\)Factor out 3:
\(S = 3[a + (p-1)d]\)From equation (1), we know that \(a + (p-1)d = k\). Substituting this value:
\(S = 3k\)Therefore, the sum of the \(p^{th}\) term, \((p+q)^{th}\) term, and \((p-q)^{th}\) term is \(3k\).
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