Question

Which term of the AP : 121,117,113,..., is its first negative term? #hint - Find n for a-n < 0

Answer 1

121,117,113.......

a=121.         d=-4

For a(n) be negative

a(n)<0

a+(n-1)d <0

121+(n-1)*(4)<0

For ans be negative

(n-1)(-4)>121

4n >(125)

n >125/4

n >31.25

n=32

Therefore the 31st term is the negative term

Verification

a(n)=a+(n-1)d

a(n)=121+(32-1)-4

a(n)=121+31*-4

a(n)=121-124

a (n)=-3

Answer 2

a = 121

d = -4

let an= 0

then

an = a + (n-1)d

=> 0 = 121 + (n-1)(-4)

=> 0 = 121 - 4n + 4

=> 4n = 125

=> n = 125/4

=> n = 31.25

as n must be natural number then n= 32

32nd term is First negative number.

a32 = 121 + (31)-4

= 121 - 124 = -3