Question

Find t_n for an Arithmetic Progression where t₃ = 22, t₁₇ = –20.

Answer 1

We know that the n^th term of an A.P. is t_n = a + (n – 1)d.
We have:
t₃ = 22
Thus, we get:
t₃ = a + (3 – 1)d = 22
$\Rightarrow$a + 2d = 22 …(1)

Also,
t₁₇ = –20
Thus, we get:
t₁₇ = a + (17 – 1)d = –20
$\Rightarrow$a + 16d = –20 …(2)

On subtracting (2) from (1), we get:
a + 2d – a – 16d = 22 – (–20)
$\Rightarrow$–14d = 42
$\Rightarrow$d = $\frac{42}{-14}=-3$

On putting d = –3 in (1), we get:
a + 2d = 22
$\Rightarrow$a + 2(–3) = 22
$\Rightarrow$a – 6 = 22
$\Rightarrow$a = 22 + 6 = 28

Thus, the n^th term of the given A.P. is:
t_n = a + (n – 1)d = 28 + (n – 1)(–3) = 28 – 3n + 3
$\Rightarrow$t_n = 31 – 3n