Which term of A.P. 121, 117, 113, ... is its first negative term? If it is the nth term, find S_n.

Formula used.

S_n = [2a + (n–1)d]

a_n = a + (n–1)d

d = a_{n + 1} – a_n

Let the 1^st negative term be X

a = 121

d = 117 – 121

= –4

If the difference is –4 then

Then X can be (–1), (–2), (–3), (–4)

a_n = a + (n–1)d

X = 121 + (n–1)(–4)

X = 125–4n

4n = 125 – X

n =

For the number of terms to be an integer (125–X) will be multiple of 4

When we put X = –1

n will be

which is not an integer

When we put X = –2

n will be

which is not an integer

When we put X = –3

n will be = 32

which is an integer

Hence –3 is the 1^st negative term.

S_n = [2a + (n–1)d]

= [2×121 + (32–1)(–4)]

= 16[242 – 124]

= 16×118

= 1888