Question

If the first, second and last term of an A.P are a, b and 2a respectively, then its sum is

Options

• \[\frac{ab}{2 (b - a)}\]

• \[\frac{ab}{b - a}\]

• \[\frac{3 ab}{2 (b - a)}\]

• none of these

Answer 1

\[\frac{3 ab}{2 (b - a)}\]

Let the A.P. be a, a+d, a+2d........a+nd.
Here, let d be the common difference and n be the total number of terms.

\[a_1 = a, \]

\[ a_2 = b\]

\[ \Rightarrow a + d = b\]

\[ \Rightarrow d = b - a . . . . . \left( 1 \right)\]

\[ a_n = 2a\]

\[ \Rightarrow a + \left( n - 1 \right)d = 2a\]

\[ \Rightarrow \left( n - 1 \right)d = a\]

\[ \Rightarrow d = \frac{a}{n - 1} . . . . . \left( 2 \right)\]

Given:

From equations \[\left( 1 \right) \text { and } \left( 2 \right),\] we have:

\[\Rightarrow \frac{a}{n - 1} = b - a\]

\[ \Rightarrow \frac{a}{b - a} + 1 = n\]

\[ \Rightarrow \frac{a + b - a}{b - a} = n\]

\[ \Rightarrow \frac{b}{b - a} = n\]

Now, sum of n terms of an A.P.:

\[S = \frac{n}{2}\left\{ a + a_n \right\}\]

\[ = \frac{n}{2}\left( 3a \right)\]

\[ = \frac{3ab}{2\left( b - a \right)}\]