Question

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Answer 1

$\underbrace{\bf{a_n=a+(n-1)d}}$

where ,

• aₙ denotes nth term's value
• a denotes first term
• n denotes nth term
• d denotes common difference

Given , 8th term is zero.

⇒ a₈ = 0

⇒ a + ( 8 - 1 ) d = 0

⇒ a + 7d = 0

⇒ a = - 7d ... (1)

We need to prove " 38th term is triple of 18 th term "

$\\\\\bf{a_{18}=a+17d}\\\\\implies \bf{a_{18}=(-7d)+17d\;[\;From\;(1)\;]}\\\\\implies \bf{\red{a_{18}=10d...(2)}}$

$\bf{a_{38}=a+37d}\\\\\implies \bf{a_{38}=(-7d)+37d\;[\;From\l(1)\;]}\\\\\implies a_{38}=30d\\\\\implies \bf{a_{38}=3(10d)}\\\\\implies \bf{\red{a_{38}=3(a_{18})\;[From\;(2)\;]}}$

Hence Proved

Answer 2

Step-by-step explanation:

Given that there is an AP.

8th term = 0

To prove that,

38th term = 3(18th term)

We know that,

nth term of an AP is given by,

a_n = a + (n-1)d

Where,

• a = first term
• d = commom difference

Therefore, we will get,

=> a_8 = a + 7d

=> a + 7d = 0

=> a = -7d

Now, we have,

=> a_38 = a + 37d

=> a_38 = -7d + 37d

= a_38 = 30d

Also, we have,

=> a_18 = a + 17d

=> a_18 = -7d + 17d

=> a_18 = 10d

=> a_18 = 30d/3

=> a_18 = a_38/3

=> a_38 = 3(a_18)

Hence proved.