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  • Show that the sum of an AP whose first term is a, the second term b and the last term c, is equal to a+c b+c-2a/2 b-a

Show that the sum of an AP whose first term is a, the second term b and the last term c, is equal to

 \frac{(a+c)(b+c-2 a)}{2(b-a)}

Answers (1)

Here

 a\textsubscript{1} = a, a\textsubscript{2} = b\\

\\ d = a\textsubscript{2} - a\textsubscript{1}\\ = b - a \\ a\textsubscript{n} = c \: \: \: \: \: \: \: \: \: \: \: \: \because a\textsubscript{n} =a + (n - 1)d \\ a\textsubscript{1} + (n - 1).d = c\: \: \: \: \: $ Put a\textsubscript{1}$ = a , d= b - a \\ (n - 1) (b - a) = c - a

\\ \mathrm{n}-1=\frac{\mathrm{c}-\mathrm{a}}{\mathrm{b}-\mathrm{a}} \\ \mathrm{n}=\frac{\mathrm{c}-\mathrm{a}}{\mathrm{b}-\mathrm{a}}+1 \\ \mathrm{n}=\frac{\mathrm{c}-\mathrm{a}+\mathrm{b}-\mathrm{a}}{\mathrm{b}-\mathrm{a}} \\ \mathrm{n}=\frac{\mathrm{c}+\mathrm{b}-2 \mathrm{a}}{\mathrm{b}-\mathrm{a}} \\ \mathrm{S}_{\mathrm{n}}=\frac{\mathrm{n}}{2}\left[2 \mathrm{a}_{1}+(\mathrm{n}-1) \mathrm{d}\right] \\ =\frac{\mathrm{c}+\mathrm{b}-2 \mathrm{a}}{2(\mathrm{~b}-\mathrm{a})}[\mathrm{a}+\mathrm{a}+(\mathrm{n}-1) \mathrm{d}] \\ =\frac{(\mathrm{c}+\mathrm{b}-2 \mathrm{a})}{2(\mathrm{~b}-\mathrm{a})}\left[\mathrm{a}+\mathrm{a}_{\mathrm{n}}\right] \quad\left[\because \mathrm{a}+(\mathrm{n}-1) \mathrm{d}=\mathrm{a}_{\mathrm{n}}\right] \\ \frac{(\mathrm{a}+\mathrm{c})(\mathrm{b}+\mathrm{c}-2 \mathrm{a})}{2(\mathrm{~b}-\mathrm{a})} \quad \quad\left[\because \mathrm{a}_{\mathrm{n}}=\mathrm{c}\right]

Hence proved.

Posted by

infoexpert21

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