Question

If the 4th term of an A.P. is 0 then prove that the 25th term is 3 times the 11th term.

Answer 1

Answer:

Step-by-step explanation:

It is given that the 4th term of the AP is 0. And, from the properties of arithmetic progressions, we know that the nth term of the AP remains a + ( n - 1 )d, where a is the first term, n is the number of terms, and d is the common difference between the terms.

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

Therefore,

= > 4th term = 0

= > a + ( 4 - 1 ) d = 0

= > a + 3d = 0

= > a = - 3d             ...( i )

So, the 11th term of the AP should be : -

= > 11th term = a + ( 11 - 1 )d

= > 11th term = a + 10d

From ( i ) , a = - 3d. So substituting the value of a,

= > 11th term = - 3d + 10d

= > 11th term = 7d

Also, the 25th term of the AP should be : -

= > 25th term = a + ( 25 - 1 )d

= > 25th term = a + 24d

From ( i ) , a = - 3d. So substituting the value of a,

= > 25th term = - 3d + 24d

= > 25th term = 21d

Hence,

= > three times of 7d = 21d

= > Three times of 11th term = 25th term

Hence proved that the the 25th term is 3 times the 11th term.

$\:$

Answer 2

Answer:

Step-by-step explanation:

Given, t4 = 0

-> a+(n-1)d = 0

a + (4-1)d = 0

a + 3d = 0

a = -3d -----(i)

T25 = a+24d

= -3d + 24d

T25 = 21d

T11 = a + 10d

T11 = -3d + 10d

T11 = 7d

Multiplying 3 on both sides

3T11 = 3 (7d)

3T11 = 21d

Therefore , 3T11 = T25 .