The 5th and 9th term of an Arithmetic Progression are 4 and -12 respectively. Find :
(a) the first term
(b) common difference
(c) sum of 16 terms of the A.P.
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If first term is a and common difference is d of the A.P.
By formula,
nth term = an = a + (n - 1)d
Given,
⇒ 5th term = 4
⇒ a5 = 4
⇒ a + (5 - 1)d = 4
⇒ a + 4d = 4 .........(1)
⇒ 9th term = -12
⇒ a9 = -12
⇒ a + (9 - 1)d = -12
⇒ a + 8d = -12 .........(2)
Subtracting equation (1) from (2), we get :
⇒ (a + 8d) - (a + 4d) = -12 - 4
⇒ a - a + 8d - 4d = -16
⇒ 4d = -16
⇒ d = $-\frac{16}{4}$
⇒ d = -4.
Substituting value of d in equation (1), we get :
⇒ a + 4d = 4
⇒ a + 4(-4) = 4
⇒ a - 16 = 4
⇒ a = 4 + 16 = 20.
(a) Hence, first term of A.P. = 20.
(b) Common difference of A.P. = -4.
(c) By formula,
Sum of n terms of A.P. = $\frac{n}{2}(a + l)$
Sum of 16 terms of A.P. = $\frac{n}{2}(a + a_{16})$
$$\begin{aligned}= \frac{16}{2}[a + a + (16 - 1)d] \\[1em] = 8(2a + 15d) \\[1em] = 8 \times (2 \times 20 + 15 \times -4) \\[1em] = 8 \times (40 - 60) \\[1em] = 8 \times -20 \\[1em] = -160.\end{aligned}$$
Hence, sum of 16 terms of A.P. = -160.