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Study Stuff for MCA Examinations - NIMCET
Study Stuff for MCA Examinations - NIMCET
Previous Year Question (PYQs)
In a sequence of $21$ terms, the first $11$ terms are in A.P. with common difference $2$ and the last $11$ terms are in G.P. with common ratio $2$. If the middle term of A.P. is equal to the middle term of G.P., then the middle term of the entire sequence is
Solution
Let $t_1$ be the first term. For the A.P.: $t_6=t_1+10$, $t_{11}=t_1+20$. For the G.P. (terms $11$ to $21$ with ratio $2$): $t_{16}=t_{11}\cdot 2^5=32t_{11}$. Given $t_6=t_{16}$: $t_1+10=32(t_1+20)\Rightarrow 31t_1=-630\Rightarrow t_1=-\dfrac{630}{31}$. Middle term of entire sequence is $t_{11}=t_1+20=-\dfrac{630}{31}+20=-\dfrac{10}{31}$.Online Test Series, Information About Examination,
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