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 and the 9th terms of an A.P. are 4 and − 8 respectively.

To find :

Which term of this A.P is zero.​

Solution :

nth term of an A.P. = a + (n - 1)d

⇒ 3rd term of A.P. = a + (3 - 1)d

⇒ 4 = a + 2d                     _(1)

Again,

⇒ 9th term = a + (9 - 1)d

⇒ -8 = a + 8d                   _(2)

Now substracting (1) from (2) :

⇒ a + 8d - a - 2d = -8 - 4

⇒ 6d = -12

⇒ d = -12/6

⇒ d = -2

∴ Common difference of A.P., d = - 2

Now putting value of 'd' in (1) :

⇒ 4 = a + 2(-2)

⇒ 4 = a - 4

⇒ a = 4 + 4

⇒ a = 8

∴ First term of A.P. = 8

Now, let the term which is 0 be nth term.

So atq,

⇒ aₙ = a + (n - 1)d

⇒ 0 = 8 + (n - 1) × (-2)

⇒ - 8 = -2n + 2

⇒ - 8 - 2 = -2n

⇒ - 10 = -2n

⇒ n = -10/-2

⇒ n = 5

∴ 5th term of A.P. = 0

Answer 2

Given:-

$\bigstar$The 3rd term of the AP is 4.

$\bigstar$The 9th term of the AP is -8.

To Find:-

$\bigstar$Which term of the AP is zero?

Solution:-

$\longrightarrow \sf a_3 = a + 2d$

$\longrightarrow \sf 4 = a + 2d$

$\longrightarrow \sf a = 4 - 2d \implies(1)$

$\bigstar$Substitute the value of a (1) in following equation.

$\longrightarrow \sf a_9 = a + 8d$

$\longrightarrow \sf a_9 = (4 - 2d)+ 8d$

$\longrightarrow \sf a_9 = 4 - 2d+ 8d$

$\longrightarrow \sf a_9 = 4 - 6d$

$\bigstar$Substitute the value of a9 in the above equation.

$\longrightarrow \sf (- 8) = 4 - 6d$

$\longrightarrow \sf (- 8 - 4)= - 6d$

$\longrightarrow \sf - 6d = - 12$

$\\ \longrightarrow \sf d = \frac{ - 12}{ 6}$

$\longrightarrow \sf d = - 2$

$\bigstar$To find the value of a Substitute the value of the d in the equation (1).

$\longrightarrow \sf a = 4 - 2d$

$\longrightarrow \sf a = 4 (- 2 \times 2)$

$\longrightarrow \sf a = 4 + 4$

$\longrightarrow \sf a = 8$

$\bigstar$To find which term of the AP is zero. We are using formula.

$\star { \overbrace{ \underbrace{ \boxed{ \boxed{ \sf{a_n = a+(n-1)d }}}}}} \star$

$\longrightarrow \sf 0 = 8+(n-1) - 2$

$\longrightarrow \sf 0 = 8+( - 2n + 2)$

$\longrightarrow \sf 2n = 8+ 2$

$\longrightarrow \sf 2n = 10$

$\\ \longrightarrow \sf n = \frac{10}{2}$

$\longrightarrow \star{ \boxed{ \underline{ \bold {n = 5}}}} \star$

$\bigstar$Hence, the 5th term of the AP will be zero.