Question

ISII
9. If the 3rd and the 9th terms of an AP are 4 and – 8 respectively, which term of this AP is
zero?​

Answer 1

Answer:

Step-by-step explanation:

3rd term = 4

a+2d = 4.......(i)[here n = 3]

[a+(n-1)d = tn]

[here n = 9]9th term = -8

a+8d = -8......(ii)

on solving (i) and (ii),

=(a+8d = -8)

- (a + 2d = 4)

...........................

  6d = 12

d = 2

substituting the value of d in (ii)

a + 8d = -8

a + 8*2 = -8

a + 16 = -8

a = -8-16

a = -24

to find the term of the AP 0,

tn = 0 [ n means the position]

tn = a+(n-1)d

a+(n-1)d = 0

-24 + (n-1)2 = 0

-24 + 2n - 2 = 0

-24-2+2n = 0

-26 + 2n = 0

2n = 26

n = 13

so the 13th term is 0.

if you want to verify then

t13 = -24 + (13-1)*2 = -24 + 12*2 = -24+24 = 0

hope it helps!

plz mark me as brainliest!

tn =