Question

If 8th term of an A.P. is zero, prove that its 38th term is triple of its 18th term.​

Answer 1

Step-by-step explanation:

Hope this is correct. Thanks for the question.

Answer 2

Given:

a₈ = 0

To prove:

a₃₈ = 3(a₁₈)

Prerequisites:

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

Where;

a ➝ First term.

n ➝ Position of the term.

d ➝ common difference.

Solution:

We've been given that:

➝ a₈ = 0

Using a$_{\sf n}$ = a + (n - 1)d we get: (where n = 8)

➝ a + (n - 1)d = 0

➝ a + (8 - 1)d = 0

➝ a + 7d = 0

➝ a = - 7d

Now, we'll have to prove that a₃₈ = 3(a₁₈).

Let's try to find the value of a₃₈ and a₁₈ separately & see if a₃₈ is triple of a₁₈

Finding the value of a₁₈.

➝ a₁₈ = a + (n - 1)d

Substitute n = 18 & a = -7d here.

➝ a₁₈ = -7d + (18 - 1)d

➝ a₁₈ = -7d + 17d

➝ a₁₈ = 10d ⇒ Relation(1)

Finding the value of a₃₈.

➝ a₃₈ = a + (n - 1)d

Substitute n = 38 & a = -7d here.

➝ a₃₈ = -7d + (38 - 1)d

➝ a₃₈ = -7d + 37d

➝ a₃₈ = 30d ⇒ Relation(2)

Now, the original statement that we had to prove was:

➝ a₃₈ = 3(a₁₈)

Substitute the value of we've got in Relation(1) and Relation(2) & we get:

➝ 30d = 3(10d)

➝ 30d = 30d

➝ LHS = RHS

Hence we've proved that the AP's 38th term is triple of its 18th term.​