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Question
The 10th and 18th terms of an A.P. are 41 and 73 respectively. Find 26th term.
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Solution
Given:
\[a_{10 =} 41\]
\[ \Rightarrow a + \left( 10 - 1 \right)d = 41 \left[ a_n = a + \left( n - 1 \right)d \right]\]
\[ \Rightarrow a + 9d = 41 \]
\[\text { And }, a_{18} = 73\]
\[ \Rightarrow a + \left( 18 - 1 \right)d = 73 \left[ a_n = a + \left( n - 1 \right)d \right]\]
\[ \Rightarrow a + 17d = 73 \]
\[\text { Solving the two equations, we get }: \]
\[ \Rightarrow 17d - 9d = 73 - 41\]
\[ \Rightarrow 8d = 32\]
\[ \Rightarrow d = 4 . . . (i)\]
\[\text { Putting the value in first equation, we get }: \]
\[a + 9 \times 4 = 41\]
\[ \Rightarrow a + 36 = 41\]
\[ \Rightarrow a = 5 . . . (ii)\]
\[a_{26} = a + \left( 26 - 1 \right)d \left[ a_n = a + \left( n - 1 \right)d \right]\]
\[ \Rightarrow a_{26} = a + 25d \]
\[ \Rightarrow a_{26} = 5 + 25 \times 4 \left( \text { From } (i) \text { and } (ii) \right)\]
\[ \Rightarrow a_{26} = 5 + 100 = 105\]
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