Question
Let $a_{1},a_{2},a_{3},\ldots\ldots$ be an A.P. and $g_{1} = a_{1},{\text{ }g}_{2},{\text{ }g}_{3},\ldots\ldots$ be an increasing G.P. If $a_{1} = a_{2} + g_{2} = 1$ and $a_{3} + g_{3} = 4$, then $a_{10} + g_{5}$ is equal to :
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Solution
Sol. [Key Concept: From $a_{1} = a_{2} + g_{2} = 1$ and $a_{3} + g_{3} = 4$ find $d$ and ratio $r$; then $a_{10} + g_{5}$.]
A.P. : $1,a_{2},a_{3},\ldots$
G.P. : $1,{\text{ }g}_{2},{\text{ }g}_{3},\ldots$
$${a_{2} + g_{2} = 1 \Rightarrow 1 + d + r = 1 \Rightarrow \text{ }d + r = 0 }{1 + 2\text{ }d + r^{2} = 4 \Rightarrow r^{2} - 2r - 3 = 0 }$$$\Rightarrow r = 3, - 1$ (reject)
$${\therefore d = - 3 }{\therefore a_{10} = 1 + 9d = 1 + 9( - 3) = - 26 }{g_{5} = 1.r^{4} = 3^{4} = 81 }{\therefore a_{10} + g_{5} = 55 }$$
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