Question

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

Answer 1

$x + 2y = 4 \\ x + 8y = - 8 \\ y = - \frac{ 5}{3} \\ x = \frac{17}{6} \\ x + (n - 1)( \frac{ - 5}{3} ) = 0 \\ \frac{17}{6 } = \frac{5}{3} (n - 1) \\ n = 2.7 \\ not \: possible$

Answer 2

Solution:

Given:

$\sf{\implies a_{3} = 4}$

$\sf{\implies a_{9} = -8}$

To Find:

=> Which term of AP is 0.

Formula used:

$\sf{\implies a_{n}=a+(n-1)d}$

So, we know that

$\sf{\implies a_{3}=a+2d}$

$\sf{\implies a+2d=4\;\;\;..........(1)}$

$\sf{\implies a_{9}=a+8d}$

$\sf{\implies a+8d=-8\;\;\;..........(2)}$

By using substitution method we will find value of a and d.

=> a + 2d = 4      .......(1)

=> a + 8d = -8    .......(2)

=> a + 2d = 4

=> a = 4 - 2d

Put the value of a in Equation (2), we get

=> a + 8d = -8

=> (4 - 2d) + 8d = -8

=> 4 + 6d = -8

=> 6d = - 8 - 4

=> 6d = -12

=> d = -12/6

=> d = -2

Put the value of d in Equation (1), we get

=> a + 2d = 4      

=> a - 4 = 4

=> a = 8

Let nth term be 0. So,

=> a + (n - 1)d = 0

=> 8 + (n - 1)-2 = 0

=> 8 -2n + 2 = 0

=> 10 - 2n = 0

=> -2n = -10

=> n = 10/2

=> n = 5 term.

So, 5th term be 0.