Get Answers to all your Questions

Answers (1)

It is given that the sum of $3\textsuperscript{rd} and 8\textsuperscript{th} term is 7$.

$\\a\textsubscript{3} = a + (3 - 1)d (using a\textsubscript{n} = a + (n-1)d)\\ a\textsubscript{3} = a + 2d\\ a\textsubscript{8} = a + (8 - 1)d (using a\textsubscript{n} = a + (n-1)d)\\ a\textsubscript{8} = a + 7d\\ a\textsubscript{3} + a\textsubscript{8 }= 7\\ a + 2d + a +7d = 7\\ 2a + 9d = 7 \ldots (1)\\ Similarly the sum of 7\textsuperscript{th} and 14\textsuperscript{th} term is - 3\\ a\textsubscript{7} + a\textsubscript{14} = - 3\\ a + (7 - 1).d + a + (14 - 1).d = - 3 (using a\textsubscript{n} = a + (n-1)d)\\ a + 6d + a + 13d = - 3\\ 2a + 19d = - 3 \ldots (2)\\$
$\\Solve (1) and (2) 2a + 19d = - 3\\ 2a + 9d = 7\\ 10d = - 10\\ d = - 1\\ Put d = - 1 in (1)\\ \Rightarrow 2a + 9(-1) = 7 \\ \Rightarrow 2a = 7 + 9 \\ \Rightarrow a=\frac{16}{2}=8 \\ a\textsubscript{10} = (a + (10 - 1)d) \\ \Rightarrow a\textsubscript{10} = 8 + 9(-1) \\ \Rightarrow a\textsubscript{10} = 8 - 9\\ \Rightarrow a\textsubscript{10} = -1\\$

View full answer