Question 18

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Jambmaths question: 

The sum of the first n term of the arithmetic progression 5,11,17,23,29,35, –, –,–

Jamb Maths Solution: 

$\begin{align}  & a=\text{ 5},\text{ }d=\text{11}\text{5}=\text{6} \\ & {{S}_{n}}=\tfrac{n}{2}[2a+(n-1)d] \\ & {{S}_{n}}=\tfrac{n}{2}[2\times 5+(n-1)6]=\tfrac{n}{2}[10+6(n-1)] \\ & {{S}_{n}}=n[5+3(n-1)]=n[5+3n-3]=n[2+3n] \\\end{align}$

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