  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  B
0  C
1  D
15  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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