How many terms are there in an A.P. whose first and fifth terms are $-14$ and $2$, respectively and the last term is $62$.

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Step 1: $a = -14$, $a_5 = 2 \Rightarrow a + 4d = 2 \Rightarrow -14 + 4d = 2 \Rightarrow d = 4$
Step 2: $a_n = 62 \Rightarrow a + (n - 1)d = 62$
Step 3: $-14 + (n - 1) 	imes 4 = 62 \Rightarrow (n - 1) 	imes 4 = 76 \Rightarrow n - 1 = 19 \Rightarrow n = 20$

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