Question

$\huge\sf\underline\purple{question}$
If the 3rd term of an AP is 4
and 9th term is -8
then which term is 0​

Answer 1

Given

${\sf a_3 = 4 ~~~~~~~ a_9 = (-8) ~~~~~a_n = ?}$

We know that

$\\ \boxed{\blue{ \: \: \: \: \: \: \sf a_n = a+(n-1)d} }$

For a_3

${ \: \: \: \: \: \: \sf \mapsto a_3 = a+(3-1)d }$

${ \: \: \: \: \: \: \sf \mapsto 4 = a+2d \: \: \: \: \: - - - - (1) }$

For a_9

${ \: \: \: \: \: \: \sf \mapsto a_9 = a+(9-1)d }$

${ \: \: \: \: \: \: \sf \mapsto - 8 = a+8d \: \: \: \: \: - - - - (2)}$

On Subtracting Equ.(2) From Equ.(1)

${ \: \: \: \: \: \: \sf \mapsto 4-(-8) = a+2d-(a+8d) }$

${ \: \: \: \: \: \: \sf \mapsto 4 + 8 = \cancel a+2d\cancel{-a} - 8d }$

${ \: \: \: \: \: \: \sf \mapsto 12 = - 6d }$

${ \: \: \: \: \: \: \sf \mapsto d = - \dfrac{12}{6} }$

${ \: \: \: \: \: \sf \mapsto d = - 2 }$

Put d in Equ.(1)

${ \: \: \: \: \: \: \sf \mapsto 4 = a + 2 \times ( - 2) }$

${ \: \: \: \: \: \: \sf \mapsto 4 = a - 4 }$

${ \: \: \: \: \: \: \sf \mapsto a = 4 + 4 }$

${ \: \: \: \: \: \: \sf \mapsto a = 8 }$

Now a = 8 and d = (-2)

To Find a_n = 0

$\\ { \: \: \: \: \: \: \sf \mapsto a_n = a+(n-1)d }$

${ \: \: \: \: \: \: \sf \mapsto 0 = 8 + (n - 1)( - 2) }$

${ \: \: \: \: \: \: \sf \mapsto 8 - 2n + 2 = 0}$

${ \: \: \: \: \: \: \sf \mapsto - 2n + 10= 0}$

${ \: \: \: \: \: \: \sf \mapsto - 2n = - 10}$

${ \: \: \: \: \: \: \sf \mapsto 2n = 10}$

${ \: \: \: \: \: \: \sf \mapsto n = 5}$

• If the 3rd term of an AP is 4 and 9th term is -8 then 5th term is 0.

◾◾◾◾◾◼️◾◼️◾◼️◾◼️◾◾◼️

Verification

$\footnotesize\sf a_n = 8+(n-1)(-2)$

$\footnotesize\sf a_n = 8+(5-1)(-2)$

$\footnotesize\sf a_n = 8+(4)(-2)$

$\footnotesize\sf a_n = 8 - 8$

$\footnotesize\sf a_n = 0$

Plz Mark At Brainllist.

Answer 2

Answer:

me to you have a great day