Question

# 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.

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 $d$.

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 $n$.