Question 20

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Jambmaths question: 

$\begin{align}  & \text{The }nth\text{ term of the progression }\tfrac{4}{2},\tfrac{7}{3},\tfrac{10}{4},\tfrac{13}{4},\cdot \cdot \cdot \text{ is} \\ & (A)\text{ }\tfrac{3n-1}{n+1}\text{ }(B)\text{ }\tfrac{1-3n}{n+1}\text{ }(C)\text{ }\tfrac{3n+1}{n+1}\text{  (D) }\tfrac{3n+1}{n-1} \\\end{align}$

Jamb Maths Solution: 

$\begin{align}  & \text{We calculate for the }{{n}^{th}}-\text{term for the numerator and denominator } \\ & \text{separately and we later merge two together} \\ & \text{For the numerator we have} \\ & \text{4, 7, 10, 13,}\cdot \cdot \cdot  \\ & {{T}_{1}}=4,{{T}_{2}}=7 \\ & \text{common difference}=d\text{ }=7-4=3 \\ & {{T}_{n}}=a+(n-1)d \\ & {{T}_{n}}=4+(n-1)3=4+3n-3=3n+1 \\ & \text{The denominator, the sequence is 2, 3, 4 5,}\cdot \cdot \cdot  \\ & {{T}_{1}}=2,\text{ }{{T}_{2}}=3,\text{ }d=3-2 \\ & {{T}_{n}}=2+(n-1)1=2+n-1=1+n \\ & \text{The }{{\text{T}}_{n}}\text{ of the sequence will be  }\frac{3n+1}{n+1} \\\end{align}$

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