Question

The 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:

a+7d=0

a=-7d......eq1

a+37d= 3(a+17d)

a+37d=3a+51d

0=2a+14d

-14d=2a

-7d=a.....eq2

From eq1 and eq2 we can say that the 38 th term is triple of its 18 th term

Answer 2

Answer:-

Given:

8th term of an AP = 0

We know that,

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

Hence,

→ a + (8 - 1)d = 0

→ a + 7d = 0

→ a = - 7d -- equation (1)

We have to prove:

38th term = 3 * 18th term.

→ a + (38 - 1)d = 3 [ a + (18 - 1)d ]

→ a + 37d = 3a + 51d

Putting the value of a from equation (1) we get,

→ - 7d + 37d = 3 ( - 7)d + 51d

→ 30d = - 21d + 51d

→ 30d = 30d

→ LHS = RHS

Hence, Proved.

Additional Information:-

• A series in which each term (except first term) differs from its preceding term by a fixed quantity is called an Arithmetic Progression (AP).

• The fixed quantity is called common difference.

• General form of an AP is a , a + d .... if a is the first term and d is the common difference.

• nth term of an AP is a + (n - 1)d