The 8 th term of an AP is zero . Prove that 38 th term is triple of its 18 th term.
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8th term = a+7d =0, so a=-7d......
18th term=a+17d =-7d +17d=10d....
38th term = a+37d ... = -7d+37d=30d....
clearly 3(10d)=30d....
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18th term=a+17d =-7d +17d=10d....
38th term = a+37d ... = -7d+37d=30d....
clearly 3(10d)=30d....
plzz mark it as a brainliest answer
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8th term of A.P = a+7d = 0
therefore a = -7d ...........(1)
now 18 th term of A.P = a+ (n-1)d = a + 17d = -7d+17d = 10d ( from ....(1) )
now 38th term = a+37d = -7d + 37d = 30d
30d is thrice of 10d hence 38th term is thrice of 18th term hence proved
therefore a = -7d ...........(1)
now 18 th term of A.P = a+ (n-1)d = a + 17d = -7d+17d = 10d ( from ....(1) )
now 38th term = a+37d = -7d + 37d = 30d
30d is thrice of 10d hence 38th term is thrice of 18th term hence proved
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