Question

The 3rd and 9th terms of an AP are 4 and – 5 respectively, which term of the AP is 0?

Answer 1

Step-by-step explanation:

a3=4

a9=-5

an=a+(n-1)d

a3=4=a+2d...(1)

a9=-5=a+8d..(2)

after solving

-6d=9

d=-9/6

d=-3/2

putting in ...(1)

a-3=4

a=7

solving it or substituting u get the answer..

Answer 2

Here's the Solution :

Let the first term and common difference of the AP are 4 and 9

Given ,

3rd term = 4

=> a + ( 3-1)d = 4

=> a + 2d = 4

=> a = 4 - 2d

Again ,

9th term = -5

=> a +(9-1)d = -5

=> a +8d = -5

=> 4 - 2d + 8d = -5 ( since a = 4- 2d)

=> 4 + 6d = -5

=> 6d = -5-4

=> 6d = -9

=> 2d =- 3

=> d =- 3/2

Now ,

a = 4 - 2×(-3/2)

=> a = 4 + 3

=> a = 7

Let us take that 0 be the nth term

So ,

0 = 7 + ( n - 1) (-3/2 )

=> -7 = ( n - 1) (-3/2 )

=> (n - 1) =-7×(-2/3 )

=> n - 1 =14/3

=> n = 1+(14/3)

=> n = 17/3

Therefore , n is a fraction , not a natural number.

Thus 0 is not a term of this AP .

Happy Learning