Show that the sum of an AP whose first term is \(a\), second term is \(b\) and last term is \(c\), is
\[ S = \dfrac{(a+c)(b+c-2a)}{2(b-a)}. \]
\(S=\dfrac{(a+c)(b+c-2a)}{2(b-a)}\)
Step 1: Recall formula for common difference.
The common difference of an AP is given by:
\(d = \text{second term} - \text{first term} = b - a.\)
Step 2: Express last term in terms of \(n\).
In an AP, the \(n^{\text{th}}\) term is: \(a_n = a + (n-1)d.\)
Here, last term \(c = a + (n-1)d.\)
So, solving for \(n\):
\(n-1 = \dfrac{c-a}{d} \;\Rightarrow\; n = \dfrac{c-a}{d} + 1.\)
Step 3: Formula for sum of \(n\) terms.
We know: \(S = \dfrac{n}{2}(\text{first term} + \text{last term}).\)
So, \(S = \dfrac{n}{2}(a+c).\)
Step 4: Substitute value of \(n\).
\(S = \dfrac{a+c}{2}\Big(\dfrac{c-a}{d} + 1\Big).\)
Step 5: Simplify.
\(S = \dfrac{a+c}{2}\cdot\dfrac{c-a+d}{d}.\)
Step 6: Replace \(d = b-a.\)
Then numerator: \(c-a+d = c-a+(b-a) = b+c-2a.\)
Final Result:
\(S = \dfrac{(a+c)(b+c-2a)}{2(b-a)}.\)
Hence proved.