In an arithmetic sequence t_2=6 and t_6=16, how do you find t_21?

Jun 19, 2016

${t}_{21} = 53 \frac{1}{2}$

Explanation:

$\textcolor{red}{\text{Preliminaries}}$

Consider the place indicator of $n$ where $n = 1$ represents the first term and $n = 2$ the second term and so on.

If $a$ is the first term then

1st term $\textcolor{w h i t e}{.} \to a$
2nd term $\to a + k$
3rd term $\to a + 2 k$
4th term $\to a + 3 k$

From this you observe than any term is $a + \left(n - 1\right) k$

Observe:

n=1 $\textcolor{w h i t e}{.} \to a + \left(1 - 1\right) k = a$
n=2$\to a + \left(2 - 1\right) k = a + k$
n=3 $\to a + \left(3 - 1\right) k = a + 2 k$
n=4 $\to a + \left(4 - 1\right) k = a + 3 k$
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so

at ${t}_{2} \to n = 2 \to a + k = 6$......Equation (1)

at ${t}_{6} \to n = 6 \to a + 5 k = 16$.....Equation (2)

Subtract equation (1) from Equation (2)

$\implies 4 k = 10$

Thus $\textcolor{b l u e}{k = \frac{10}{4} = \frac{5}{2}}$...........................(3) '~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Substitute (3) back into Equation (1)

color(brown)(a+k=6)color(blue)(->a+5/2=6

Subtract $\frac{5}{2}$ from both sides

$\textcolor{b l u e}{a = \frac{7}{2}}$
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Check:
${t}_{2} = 6 \to \frac{7}{2} + \frac{5}{2} = 6$ as required
${t}_{6} = 16 \to \frac{7}{2} + \left(5 \times \frac{5}{2}\right) = 16$ as required
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$\textcolor{red}{\text{Answering the questions}}$

color(brown)(t_21 = a+20k)color(blue)(->7/2+(20xx5/2 )=53 1/2