Question

if the 3rd and 9th term of an ap are 4 and -8 respectively which term of this ap is 0

Answer 1

-------------------------------

♣ Given :-

For an A.P :

• 3rd Term = $\sf a_3$ = 4

• 9th term = $\sf a_9$ = - 8

-------------------------------

♣ To Find :-

• Which term of the corresponding A.P is 0

-------------------------------

♣ Main Formula :-

nth term of an A.P is given by formula :

$\bf \large a_n = a + (n - 1)d$

Where :

• a = First Term

• d = Common Difference

-------------------------------

♣ Solution :-

Let a and d be the first term and common difference of the corresponding A.P. respectively.

Hence ,

$\large a_3 = 4 \\ \\ \large\pmb{ a + 2d = 4} \: \: \: \: - - - - (1)$

$\large a_9 = - 8 \\ \\ \large\pmb{ a + 8d = - 8} \: \: \: \: - - - - (2)$

$\Large \bigstar \:   \underline{ \pmb{ \mathfrak{ \text{S}ubtracting \: (1) \: From \: (2) }}}$

$:\longmapsto6d = - 12 \\ \\ \LARGE \purple{ :\longmapsto  \underline {\boxed{{\bf d = - 2} }}}$

$\Large \bigstar \: \underline{ \pmb{ \mathfrak{ Putting \: \text{V}alue \: of \: \text{d} \: in \: (1) }}}$

$a + 2 \times ( - 2) = 4 \\ \\ \LARGE\purple{ :\longmapsto  \underline {\pmb{{\boxed{{a = 8} }}}}}$

Now,

Let nth term of the A.P. is zero :

Hence ,

$a_n = 0 \\ \\ :\longmapsto a + (n - 1)d = 0 \\ \\ :\longmapsto 8 + (n - 1)( - 2) = 0 \\ \\ :\longmapsto8 - 2n + 2 = 0 \\ \\ :\longmapsto2n = 10 \\ \\ \LARGE \purple{ :\longmapsto  \underline {\boxed{{\bf n = 5} }}}$

$\large \underline{ \pink{\pmb{ \mathfrak{ \text{H}ence \: 5th \: term \: of \: the \: \bold{ A.P} \: i s \: \bf 0}}}}$

$\Large \red{\mathfrak{  \text{W}hich \:   \: is  \:  \: the  \:  \: required} }\\ \huge \red{\mathfrak{ \text{ A}nswer.}}$

-------------------------------

Answer 2

5th term

Step-by-step explanation:

QUESTION :-

If the 3rd term and 9th term of an A.P. are 4 and -8 respectively, which term of this A.P. is 0.

_______________________

SOLUTION :-

We know that,

nth term of A.P.

$\bf {a}_{n}$ = a + (n - 1)d

[where a = first term, n = nth term & d = common difference]

_______________________

So,

3rd term = 4

$\bf {a}_{3}$ = a + (3 - 1)d = 4

= a + 2d = 4.....(i)

9th term = -8

$\bf {a}_{9}$ = a + (9 - 1)d = -8

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

_____________________

By subtracting (i) from (ii)

a + 2d = 4

- a ± 8d = ±8

=> -6d = 12

=> d = 12/-6

=> d = -2

Now put the value of d in eq. (i)

a + 2d = 4

=> a + 2(-2) = 4

=> a - 4 = 4

=> a = 4+4

=> a = 8

_____________________

Question is which term of A.P. is 0, let the nth term of A.P. = 0

=> $\bf {a}_{n}$ = 0

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

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

=> 8 - 2n + 2 = 0

=> 8 + 2 - 2n = 0

=> 10 - 2n = 0

=> - 2n = -10

=> n = -10/-2

=> n = 5

So,

5th term of A.P. is 0.

______________________

Hope it helps.

#BeBrainly