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Since this sum involves finding the value of $ {N^{th}} $ term , we will have to use the formula $ {T_n} = a + (n - 1)d $ , where

$ {T_n} $ - Value of $ {N^{th}} $ term

$ a $ - $ 1st $ term of the given series

$ n $ - No. of terms in the Series

$ d $ - common difference between $ 2 $ consecutive terms in the series

Considering the first statement of the numerical, we can form the following equations

$ {T_5} = a + 4d...............(1) $

$ {T_2} = a + d.................(2) $

Given that $ {T_5} = 3 \times {T_2} $

From $ Equation1\& Equation2 $ we can form the following relation

$ \Rightarrow a + 4d = 3 \times (a + d)...........(3) $

$ \Rightarrow a + 4d = 3a + 3d...........(4) $

$ \Rightarrow d = 2a...........(5) $

Considering second statement in the numerical ,we can form following equations

$ {T_{12}} = a + 11d............(6) $

$ {T_6} = a + 5d............(7) $

It is given that $ {T_{12}} - {T_6} = 1.............(8) $

Substituting values of $ Equation6\ & \ Equation7 $ in $ Equation8 $ we get the following equation

$ \Rightarrow (a + 11d) - (a + 5d) = 1 $

\[ \Rightarrow 11d - 5d = 1\]

\[ \Rightarrow d = \dfrac{1}{6}\]

We can use $ Equation5 $ to find the value of $ a $ .

$ \Rightarrow a = \dfrac{1}{6} \times 2 = \dfrac{1}{3} $

Since we have the value of $ a\& d $ we can now find the $ 16th $ term

$ \Rightarrow {T_{16}} = a + 15d $

Substituting values of $ a\& d $ we get the final value of $ 16th $ term

$ \Rightarrow {T_{16}} = \dfrac{1}{3} + 15 \times \dfrac{1}{6} $

In order to simplify the sum we can multiply the numerator and denominator of $ \dfrac{1}{3} $ by $ 2 $ , so that we have a common denominator.

$ \Rightarrow {T_{16}} = \dfrac{2}{6} + \dfrac{{15}}{6} $

$ \Rightarrow {T_{16}} = \dfrac{{17}}{6} $

Thus the final Answer is $ {T_{16}} = \dfrac{{17}}{6} $

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