When we divide the ninth term of an arithmetic progression by its second term, we get 5 as quotient

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When we divide the ninth term of an arithmetic progression by its second term, we get 5 as quotient and when we divide the thirteenth term of that progression by the sixth term, we get 2 as a quotient and 5 as a remainder, then

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Sol. [Key Concept: A.P.: the ninth-by-second-term division fixes first term & common difference.]$$a_{9} = 5a_{2}\ \Rightarrow \ a_{1} + 8	ext{ }d = 5\left( a_{1} + d ight)\ \Rightarrow 4a_{1} = 3	ext{ }d$$$$\begin{matrix} a_{13} = 2a_{6} + 5 & \ \Rightarrow & \ \ \ \ \ \ \ \ & a_{1} + 12d = 2\left( a_{1} + 5d ight) + 5 \Rightarrow a_{1} + 5 = 2d \ & \ \Rightarrow & \ \ \ \ \ \ \ \ & a_{1} = 3,d = 4 \end{matrix}$$✔ (a) $4a_{1 = 3}d,\ 2d - a_{1 = 5} \Rightarrow \ a_{1 = 3}$✔ (b) $a_{7 = 3 + 6(4) = 27}$✘ (c) $a_{1 = 3} 
eq 4$✘ (d) third term $= 11$

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