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Question
The 6th and 17th terms of an A.P. are 19 and 41 respectively, find the 40th term.
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Solution
Given:
\[a_6 = 19\]
\[ \Rightarrow a + \left( 6 - 1 \right)d = 19 \left[ a_n = a + \left( n - 1 \right)d \right]\]
\[ \Rightarrow a + 5d = 19 . . \left( 1 \right)\]
\[\text { And,} a_{17} = 41\]
\[ \Rightarrow a + \left( 17 - 1 \right)d = 41 \left[ a_n = a + \left( n - 1 \right)d \right]\]
\[ \Rightarrow a + 16d = 41 . . \left( 2 \right)\]
\[\text { Solving the two equations, we get, } \]
\[16d - 5d = 41 - 19\]
\[ \Rightarrow 11d = 22\]
\[ \Rightarrow d = 2 \]
\[\text { Putting d }= 2 \text { in the eqn } \left( 1 \right), \text { we get }: \]
\[a + 5 \times 2 = 19\]
\[ \Rightarrow a = 19 - 10\]
\[ \Rightarrow a = 9 \]
We know:
\[a_{40} = a + \left( 40 - 1 \right)d \left[ a_n = a + \left( n - 1 \right)d \right]\]
\[ = a + 39d\]
\[ = 9 + 39 \times 2 \]
\[ = 9 + 78 = 87\]
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