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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
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Solution
\[\frac{3 ab}{2 (b - a)}\]
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.:
\[S = \frac{n}{2}\left\{ a + a_n \right\}\]
\[ = \frac{n}{2}\left( 3a \right)\]
\[ = \frac{3ab}{2\left( b - a \right)}\]
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