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The 8th term of an AP is zero.Prove that its 38th term is triple its 18th term .
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2
a(8) = a+7d, where a is the first term and d is the common difference.
a+7d=0
a=-7d
a(38) = a+37d = -7d+37d = 30d
a(18)=a+17d = -7d+17d = 10d
30 d = 3 times 10 d
Therefore, its 38th term is triple its 18th term.
a+7d=0
a=-7d
a(38) = a+37d = -7d+37d = 30d
a(18)=a+17d = -7d+17d = 10d
30 d = 3 times 10 d
Therefore, its 38th term is triple its 18th term.
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14
a+7d=0
a=-7d
a38=a+37d=-7d+37d=30d .....(i)
a18=a+17d=-7d+17d=10d ...(ii)
frm (i) and (ii)
hence a38=3a18
a=-7d
a38=a+37d=-7d+37d=30d .....(i)
a18=a+17d=-7d+17d=10d ...(ii)
frm (i) and (ii)
hence a38=3a18
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