Question

The 8th term of an AP is 0 . Prove that its 38th term is triple of its 18th term.

Answer 1

Step-by-step explanation:

Let the common difference be 'd' .

Let the common difference be 'd' .given that a+7d=0 i.e. eight term is zero

Let the common difference be 'd' .given that a+7d=0 i.e. eight term is zero we get a= -7d

Let the common difference be 'd' .given that a+7d=0 i.e. eight term is zero we get a= -7d we need to prove that {a+37d} = 3{a+17d} .

Let the common difference be 'd' .given that a+7d=0 i.e. eight term is zero we get a= -7d we need to prove that {a+37d} = 3{a+17d} .After putting the value of a as -7 we get the following equation

Let the common difference be 'd' .given that a+7d=0 i.e. eight term is zero we get a= -7d we need to prove that {a+37d} = 3{a+17d} .After putting the value of a as -7 we get the following equation from LHS we have {-7d+37d} = 30d

Let the common difference be 'd' .given that a+7d=0 i.e. eight term is zero we get a= -7d we need to prove that {a+37d} = 3{a+17d} .After putting the value of a as -7 we get the following equation from LHS we have {-7d+37d} = 30dAnd from RHS we have,

Let the common difference be 'd' .given that a+7d=0 i.e. eight term is zero we get a= -7d we need to prove that {a+37d} = 3{a+17d} .After putting the value of a as -7 we get the following equation from LHS we have {-7d+37d} = 30dAnd from RHS we have, {3(-7d+17d)}

Let the common difference be 'd' .given that a+7d=0 i.e. eight term is zero we get a= -7d we need to prove that {a+37d} = 3{a+17d} .After putting the value of a as -7 we get the following equation from LHS we have {-7d+37d} = 30dAnd from RHS we have, {3(-7d+17d)} = 3(10d)

Let the common difference be 'd' .given that a+7d=0 i.e. eight term is zero we get a= -7d we need to prove that {a+37d} = 3{a+17d} .After putting the value of a as -7 we get the following equation from LHS we have {-7d+37d} = 30dAnd from RHS we have, {3(-7d+17d)} = 3(10d)=30d = RHS

From above we get LHS=RHS =30d

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