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  • The 26th, 11thand the last term of an AP are 0, 3 and –1/5 , respectively. Find the common difference and the number of terms.

The 26th, 11thand the last term of an AP are 0, 3 and -1/5 , respectively. Find the common difference and the number of terms.

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\\$Given a\textsubscript{26}$ = 0, a\textsubscript{11} = 3, a\textsubscript{n} = - \frac{1}{5} \\ a + 25d = 0 $ \ldots $ .(1) ($using a\textsubscript{n} $= a + (n-1)d)\\

\\a + 10d +3 $ \ldots $ .(2)(using a\textsubscript{n} = a + (n-1)d)\\ \\ 15d = - 3\\

\\$Put d$ = -\frac{1}{5} $in (1) we get\\$ a - 5 = 0\\ a = 5\\ a\textsubscript{n }=\textsubscript{ } -\frac{1}{5} \\

$We know that a\textsubscript{n}$ = a + (n - 1)d\\

\\\Rightarrow -\frac{1}{5}-5=\left( n-1 \right)\left( -\frac{1}{5} \right) \\ \Rightarrow \frac{-1-25}{5}=\left( n-1 \right)\left( -\frac{1}{5} \right) \\ \Rightarrow \frac{-26}{5}=\left( n-1 \right)\left( -\frac{1}{5} \right) \\ \Rightarrow \frac{-26}{5}\times \frac{-5}{1}=n-1 \\ \therefore 26 = n - 1\\ \Rightarrow n = 26 + 1\\ \Rightarrow n = 27 \\ ANS;- \\ d= -\frac{1}{5} , n = 27\\

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