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Question

The sum of first $$16$$ terms of an $$A.P.$$ is $$120$$. Find the second term of another $$A.P.$$ whose first term is $$\dfrac{2}{15}$$ times of the first term of given $$A.P.$$ and common differences same as the common difference of given $$A.P.$$.


A
15
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B
0
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C
1
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D
15
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Solution

The correct option is C $$1$$
Let  $${ a }_{ n\quad  }$$$${ b }_{ n }$$     $$d$$ represent the nth terms and the common difference of the two given APs 

Given:
$${ S }_{ 16 }=120\\$$ 
$$\\ { b }_{ 1 }=\dfrac { 2 }{ 15 } { a }_{ 1 }$$ 

Applying the formula
$${ S }_{ n }=\dfrac { n }{ 2 } \left( 2a+(n-1)d \right) $$

We get
$${ S }_{ 16 }=\dfrac { 16 }{ 2 } \left( 2a+(16-1)d \right) =120\\$$
$$ 8(2a+15d)=120\\$$
$$ 2a+15d=15$$

Dividing the equation by 15 on both sides we get-:
$$\dfrac { 2 }{ 15 } { a }_{ 1 }+d=1\\ Using\quad { b }_{ 1 }=\dfrac { 2 }{ 15 } { a }_{ 1 }\\$$
$$ We\quad get\quad { b }_{ 1 }+d=1\\$$
$$As\quad { b }_{ 2 }={ b }_{ 1 }+d\\$$
$$ { b }_{ 2 }=1$$

Hence option (C) is the correct option.

Mathematics

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