1.

The 8th term of an A.P. is zero. Prove that its 38thterm is triple of its 18th term.​

Answer»

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


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