Question

y find the next two
given
below. 5, 8, 11, 1h..
teams of
an AP​

Answer 1

Question:

To find the next two terms of the AP 5, 8, 11 . . . . .

Explanation:

To find the next two terms of the given AP, we need to know the first term (a), the common difference (d) of the AP.

First term (a):

⇒ The first term (a) = 5

Common difference:

⇒ a₂ - a₁ = a₃ - a₂

⇒ 8 - 5 = 11 - 8

⇒ 3 = 3

∴ The common difference (d) = 3.

We know that a term of an AP, let's say 'n', can be expressed of the form;

$\sf \Longrightarrow a_n = a + (n - 1)d$

Where;

a$_{\sf n}$ = Position of the term.

a = First term.

n = Value of the term.

d = Common difference

Now, we'll find the 4th & 5th term of the AP

First, we find the 4th term:

$\sf \Longrightarrow a_n = a + (n - 1)d$

Where n = 4

$\sf \Longrightarrow a_4 = 5 + (4 - 1)3$

$\sf \Longrightarrow a_4 = 5 + (3)3$

$\sf \Longrightarrow a_4 = 5 + 9$

$\sf \Longrightarrow a_4 = 14$

Now, we find the 5th term:

$\sf \Longrightarrow a_n = a + (n - 1)d$

Where n = 5

$\sf \Longrightarrow a_5 = 5 + (5 - 1)3$

$\sf \Longrightarrow a_5 = 5 + (4)3$

$\sf \Longrightarrow a_5 = 5 + 12$

$\sf \Longrightarrow a_5 = 17$

Now, we find the 6th term:

(I'm finding the 6th term as well since the terms given in the question is incomplete I guess)

$\sf \Longrightarrow a_n = a + (n - 1)d$

Where n = 6

$\sf \Longrightarrow a_6 = 5 + (6 - 1)3$

$\sf \Longrightarrow a_6 = 5 + (5)3$

$\sf \Longrightarrow a_6 = 5 + 15$

$\sf \Longrightarrow a_6 = 20$

Final answers:

4th term = 14

5th term = 17

6th term = 20

Answer 2

★Question:–

Find the next 2 AP (5,8,11).

★To find :–

4th and 5th term.

★Solution:–

★Formula used:

$a_n = a + (n - 1)d$

$a_n = terms \: position \:$

→a=first term

→n=nth term

→d=common difference

Now,

Given AP

5,8,11

$a_1 = 5 \\ a_2 = 8 \\ a_3 = 11 \\ d =a_2 - a_1 = 8 - 5 = 3 \\ d = a_3 - a_2 = 11 - 8 = 3$

common difference (d)=3

first term (a)=5

Now,

4 th term:–

$a_n = a + (n - 1)d \\a_4 = 5 + (4 - 1)3 \\ a_4 = 5 + 9 \\ a_4 = 14$

5 th term:–

$a_n = a + (n - 1)d \\ a_5 = 5 + (5 - 1)3 \\ a_5 = 5 + 12 \\ a_5 = 17$

6 th term :–

$a_n = a + (n - 1)d \\a_6 = 5 + (6 - 1)3 \\a_6 = 5 + 15 \\ a_6 = 20$

7 th term:–

$a_n = a + (n - 1)d \\ a_7 = 5 + (7 - 1)3 \\ a_7 = 5 + 18 \\ a_7 = 23$

therefore,

$a_4 = 14 \\ a_5 = 17 \\ a_6 = 20 \\a_7 = 23$