Question

The fourth term an AP is 0.Prove that its 25th is triple its 11th term.

Answer 1

Answer:-

Given:

Forth term of an AP = 0

We know that,

nth term of an AP = a + (n - 1)d

→ a(4) = a + 3d

→ a + 3d = 0 -- equation (1).

We have to prove that it's 25th term is triple it's 11th term.

→ a(25) = 3*a(11)

→ a + (25 - 1)d = 3(a + 10d)

→ a + 24d = 3a + 30d

→ a + 24d - 30d = 3a

→ a - 6d = 3a

→ 3a - a + 6d = 0

→ 2a + 6d = 0

→ 2(a + 3d) = 0

Substitute the value of "(a + 3d)" from equation (1).

→ 2(0) = 0

→ 0 = 0

Hence ,Proved.

Answer 2

$\huge \rm \fbox{Solution :)}$

Let , the first term and common difference of AP be a and d

Given that ,

The fourth term of AP is 0 and 25th term is triple of its 11th term

Thus ,

a + (4 - 1)d = 0

a + 3d = 0

And

a + (25 - 1)d = 3{a + (11 - 1)d}

a + 24d = 3{a + 10d}

a + 24d = 3a + 30d

2a + 6d = 0

2(a + 3d) = 0

2(0) = 0 { because , a + 3d = 0 }

0 = 0

Hence , proved