nth Term of Linear Sequences

nth Term of Linear Sequences

Deducing the \(n\)th term formula of a linear (arithmetic) sequence is one of the most frequently tested algebra skills. The formula expresses the value of any term directly from its position number \(n\), without needing to list previous terms. The method is systematic and reliable.

The Method

For any arithmetic sequence:

  1. Find the common difference \(d\): subtract any term from the next. If the sequence is given as a list, compute \(T_2-T_1\).
  2. Write the formula skeleton: \( T_n = dn + c \).
  3. Find \(c\): substitute \(n=1\) and set equal to the first term: \( c = T_1 - d \).
  4. Verify: check that \(T_n = dn+c\) gives the correct value for at least one other term.

Interpreting the Formula

The nth term formula \(T_n = dn+c\) is a linear expression in \(n\). Plotting term value against position number gives a straight line with gradient \(d\) and y-intercept \(c\). The formula links sequences directly to linear graphs.

Harder Cases

Sometimes the sequence is given with a non-unit starting position, or terms are given as algebraic expressions. In these cases:

  • If given \(T_p\) and \(T_q\), find \(d = (T_q-T_p)/(q-p)\), then find \(c = T_p - dp\).
  • If terms are given as expressions in a parameter, set up equations using the equal-difference property and solve.

Worked Examples

Find the \(n\)th term of 6, 11, 16, 21, 26, …

\(d=5\). Skeleton: \(T_n=5n+c\). At \(n=1\): \(5+c=6 Rightarrow c=1\).

\[ T_n = 5n+1 \]

Check: \(T_3=16 Rightarrow 5(3)+1=16 ✓\)

Find the \(n\)th term of 20, 15, 10, 5, 0, …

\(d=-5\). \(T_n=-5n+c\). At \(n=1\): \(-5+c=20 Rightarrow c=25\).

\[ T_n = -5n+25 = 5(5-n) \]
Find the \(n\)th term of \( \frac{1}{2}, 1, \frac{3}{2}, 2, \frac{5}{2}, ldots \)

\(d=\frac{1}{2}\). \(T_n=\frac{1}{2}n+c\). At \(n=1\): \(\frac{1}{2}+c=\frac{1}{2} Rightarrow c=0\).

\[ T_n = \frac{n}{2} \]
Find the \(n\)th term of 0.3, 0.9, 1.5, 2.1, …

\(d=0.6\). \(T_n=0.6n+c\). At \(n=1\): \(0.6+c=0.3 Rightarrow c=-0.3\).

\[ T_n = 0.6n - 0.3 = 0.3(2n-1) \]
The 4th term of an arithmetic sequence is 19 and the 7th term is 31. Find the \(n\)th term.
\[ d = \frac{31-19}{7-4} = \frac{12}{3} = 4 \]

\(T_n=4n+c\). Using \(T_4=19\): \(16+c=19 Rightarrow c=3\).

\[ T_n = 4n+3 \]

Verify: \(T_7=28+3=31 ✓\)

 Key Takeaways

  • Arithmetic nth term: \(T_n=dn+c\). \(d\) is the common difference, \(c=T_1-d\).
  • The common difference is the coefficient of \(n\) — it acts exactly like the gradient in a linear equation.
  • To find \(d\) from two given terms at positions \(p\) and \(q\): \( d=(T_q-T_p)/(q-p) \).
  • Always verify by substituting a term other than the one used to find \(c\).
  • The formula works for non-integer differences just as well — e.g. \(d=\frac{1}{2}\) or \(d=0.6\).