Question

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

Answer 1

A.P. = a4 = 0

So , a + ( 4 - 1) d =0

a + 3d = 0

a = -3d .....(1)

25th term = a25

= a + (25 - 1) d

= -3d + 24d ..... ( from 1st one)

= 21 d

3 times 11 th term of an A.P. = 3a11

= 3 [ a + (11 - 1) d]

3 [ a + 10 d]

3 [ - 3d + 10d ]

3 × 7d

= 21 d

So , a25 = 3a11

Answer 2

$\huge\mathfrak\orange{Question}$

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

$\huge\mathfrak\red{Answer}$

let first term of AP is a and common difference b

4th term = a + (4 - 1)

= d = a + 3d = 0

a = 3d

Now,

T25 = a + (25-1) d = a + 24d 3d + 24d = 21d

T11 = a + (11 - 1) d = a + 10d = 3d + 10d = 7d

here it is clear that,

T25 = 3 × t11