Question

If the 3rd and the 9th terms of an A.P. are 4 and − 8 respectively, then which term of this A.P is zero.​

Answer 1

GIVEN :-

• 3rd term of A.P is 4.

• 9th term of A.P is -8.

TO FIND :-

• The term when the A.P is 0.

FORMULA USED :-

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

Here ,

• a(n) → 'n'th term
• a → First term
• n → Total number of terms
• d → Common difference

SOLUTION :-

♦ 3rd term of A.P is 4.

We know , a(n) = a + (n-1)d

Here ,

• a(n) = 4
• n = 3

Putting values we get..

$\implies\sf4 = a + (3 - 1)d$

$\\ \implies \sf 4 = a + 2d \: \: \: \: \: \: \: \: \: \: \: \: - - - (i)$

♦ 9th term of A.P is -8.

Now with the same formula ...we get ,

$\\ \implies \sf - 8 = a + (9 - 1)d$

$\\ \implies \sf \: - 8 = a + 8d \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: - - - (ii)$

Now , we will subtract equation (i) by equation (ii).

We get...

$\\ \to \sf \: - 8 - 4 = a + 8d - (a + 2d)$

$\\ \to \sf \: - 12 = a + 8d - a - 2d$

$\\ \to\sf \: - 12 = 6d$

$\\ \to \boxed{\sf \: d = - 2}$

Now , we will substitute the value of "d" in equation (i).

$\\ \underline{ \sf \: 4 = a + 2d}$

$\\ \to \sf \: 4 = a + 2( - 2)$

$\\ \to \sf \: 4 = a - 4$

$\\ \to\boxed{ \sf \:a = 8 }$

Let a(n) of the A.P be 0.

$\\ \implies\sf \: 0 = a + (n - 1)d$

Substituting values of 'a' and 'b' as 8 and -2 respectively , we get...

$\\ \implies \sf0 = 8 + (n - 1)( - 2)$

$\\ \implies\sf \: - 8 = (n - 1)( - 2)$

$\\ \implies \sf - 8 = - 2n + 2$

$\\ \implies\sf2n = 2 + 8$

$\\ \implies\sf 2n = 10$

$\\ \implies \boxed{\boxed{ \bf \:n = 5 }}$

Hence , 5th term of the A.P is 0.

Answer 2

Answer:

Given that,

3rd term, a3 = 4

and 9th term, a9 = −8

We know that, the nth term of AP is;

an = a + (n − 1) d

Therefore,

a3 = a + (3 − 1) d

4 = a + 2d ……………………………………… (i)

a9 = a + (9 − 1) d

−8 = a + 8d ………………………………………………… (ii)

On subtracting equation (i) from (ii), we will get here,

−12 = 6d

d = −2

From equation (i), we can write,

4 = a + 2 (−2)

4 = a − 4

a = 8

Let nth term of this A.P. be zero.

an = a + (n − 1) d

0 = 8 + (n − 1) (−2)

0 = 8 − 2n + 2

2n = 10

n = 5

Hence, 5th term of this A.P. is 0.