Question

The 10th and 18th terms of an A.P. are 41 and 73 respectively. Find 26th term.

Answer 1

Given:

\[a_{10 =} 41\]

\[ \Rightarrow a + \left( 10 - 1 \right)d = 41 \left[ a_n = a + \left( n - 1 \right)d \right]\]

\[ \Rightarrow a + 9d = 41 \]

\[\text { And }, a_{18} = 73\]

\[ \Rightarrow a + \left( 18 - 1 \right)d = 73 \left[ a_n = a + \left( n - 1 \right)d \right]\]

\[ \Rightarrow a + 17d = 73 \]

\[\text { Solving the two equations, we get }: \]

\[ \Rightarrow 17d - 9d = 73 - 41\]

\[ \Rightarrow 8d = 32\]

\[ \Rightarrow d = 4 . . . (i)\]

\[\text { Putting the value in first equation, we get }: \]

\[a + 9 \times 4 = 41\]

\[ \Rightarrow a + 36 = 41\]

\[ \Rightarrow a = 5 . . . (ii)\]

\[a_{26} = a + \left( 26 - 1 \right)d \left[ a_n = a + \left( n - 1 \right)d \right]\]

\[ \Rightarrow a_{26} = a + 25d \]

\[ \Rightarrow a_{26} = 5 + 25 \times 4 \left( \text { From } (i) \text { and } (ii) \right)\]

\[ \Rightarrow a_{26} = 5 + 100 = 105\]