If $$\frac{1}{{x + 2}},$$ $$\frac{1}{{x + 3}},$$ $$\frac{1}{{x + 5}}$$ are in A.P. then x = ?
A. 5
B. 3
C. 1
D. 2
Answer: Option C
Solution(By Examveda Team)
$$\eqalign{ & \frac{1}{{x + 2}},\,\frac{1}{{x + 3}},\,\frac{1}{{x + 5}}\,{\text{are}}\,{\text{in}}\,{\text{A}}{\text{.P}}{\text{.}} \cr & \therefore \frac{1}{{x + 3}} - \frac{1}{{x + 2}} = \,\frac{1}{{x + 5}} - \frac{1}{{x + 3}} \cr & \Rightarrow \frac{{x + 2 - x - 3}}{{\left( {x + 3} \right)\left( {x + 2} \right)}} = \frac{{x + 3 - x - 5}}{{\left( {x + 5} \right)\left( {x + 3} \right)}} \cr & \Rightarrow \frac{{ - 1}}{{\left( {x + 3} \right)\left( {x + 2} \right)}} = \frac{{ - 2}}{{\left( {x + 5} \right)\left( {x + 3} \right)}} \cr & \Rightarrow \frac{{ - 1}}{{x + 2}} = \frac{{ - 2}}{{x + 5}} \cr & \Rightarrow - 2x - 4 = - x - 5 \cr & \Rightarrow - 2x + x = - 5 + 4 \cr & \Rightarrow - x = - 1 \cr & \therefore x = 1 \cr} $$Related Questions on Progressions
Find the first term of an AP whose 8th and 12th terms are respectively 39 and 59.
A. 5
B. 6
C. 4
D. 3
E. 7
The sum of the first 16 terms of an AP whose first term and third term are 5 and 15 respectively is
A. 600
B. 765
C. 640
D. 680
E. 690
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