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The 17th term of an A.P. is 5 more than twice its 8th term. If the 11th term of the A.P. is 43, find the nth term.

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Hint: Use the general (nth) term of A.P. which is ${T_n} = a + \left( {n - 1} \right)d$. Satisfy the conditions given in the question and find the value of $a$ and $d$.

Complete step-by-step answer:
We know that the general (nth) term of A.P. is ${T_n} = a + \left( {n - 1} \right)d$ where $a$is the first term and $d$is the common difference.

And according to the question, the 17th term of an A.P. is 5 more than twice its 8th term. So, we have:
$ \Rightarrow {T_{17}} = 2{T_8} + 5$

Using the formula of ${T_n}$, we’ll get:
$
   \Rightarrow a + \left( {17 - 1} \right)d = 2\left[ {a + \left( {8 - 1} \right)d} \right] + 5, \\
   \Rightarrow a + 16d = 2a + 14d + 5, \\
   \Rightarrow a - 2d = - 5 .....(i) \\
$
Further, it is given that the 11th term of the A.P. is 43. So, we have:
$
   \Rightarrow a + \left( {11 - 1} \right)d = 43, \\
   \Rightarrow a + 10d = 43 .....(ii) \\
$
Now, subtracting equation $(ii)$ from equation $(i)$ we’ll get:
$
   \Rightarrow a - 2d - a - 10d = - 5 - 43, \\
   \Rightarrow - 12d = - 48, \\
   \Rightarrow d = 4 \\
$
Putting the value of d in equation $(i)$, we’ll get:
$
   \Rightarrow a - 2 \times \left( 4 \right) = - 5, \\
   \Rightarrow a - 8 = - 5, \\
   \Rightarrow a = 3 \\
$
Putting values of $a$ and $d$in general equation, we’ll get:
$
   \Rightarrow {T_n} = a + \left( {n - 1} \right)d, \\
   \Rightarrow {T_n} = 3 + \left( {n - 1} \right) \times 4, \\
   \Rightarrow {T_n} = 3 + 4n - 4, \\
$
$ \Rightarrow {T_n} = 4n - 1$

Thus, the nth term of A.P. is $4n - 1$.

Note: The general term of an A.P. is always a 1 degree polynomial in $n$ while the sum of first $n$ terms on the A.P. is a 2 degree polynomial in $n$.
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