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The 6th and 17th Terms of an A.P. Are 19 and 41 Respectively, Find the 40th Term. - Mathematics

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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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Chapter 19: Arithmetic Progression - Exercise 19.2 [Page 12]

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RD Sharma Mathematics [English] Class 11
Chapter 19 Arithmetic Progression
Exercise 19.2 | Q 8 | Page 12

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