In an $A . P$ if $a = 2$ and $t_{5} = 12$ , find $t_{2}, t_{3}, t_{4}$ .

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Solution

$a = 2 t_{5} = 12 \Rightarrow a + 4 d = 12 \Rightarrow 2 + 4 d = 12 \Rightarrow 4 d = 12 - 2 \Rightarrow 4 d = 10 \Rightarrow d = \frac{10}{4} ∴ d = \frac{5}{2} t_{2} = a + d = 2 + \frac{5}{2} = \frac{4 + 5}{2} = \frac{9}{2} = 4.5 t_{3} = a + 2 d = 2 + 2 \times \frac{5}{2} = 2 + 5 = 7 t_{4} = a + 3 d = 2 + 3 \times \frac{5}{2} = 2 + \frac{15}{2} = \frac{4 + 15}{2} = \frac{19}{2} = 9.5 t_{2} = 4.5 o r 4 \frac{1}{2} t_{3} = 7 t_{4} = 9.5 o r 9 \frac{1}{2}$

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