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JEE Main & Advanced JEE Main Paper (Held on 10-4-2019 Afternoon)

  • question_answer
    Let a, b and c be in G. P. with common ratio where\[a\ne 0\]and \[0<r\le \frac{1}{2}.\] If 3a, 7b and 15c are the first three terms of an A. P., then the 4th term of this A. P. is: [JEE Main 10-4-2019 Afternoon]

    A) \[\frac{7}{3}a\]                                  

    B) \[a\]

    C) \[\frac{2}{3}a\]                                  

    D) \[5a\]

    Correct Answer: B

    Solution :

    \[b=ar\]           \[c=a{{r}^{2}}\] 3a, 7b and 15 c are in A.P. \[\Rightarrow 14b=3a+15c\] \[\Rightarrow 14(ar)=3a+15a{{r}^{2}}\] \[\Rightarrow 14r=3+15{{r}^{2}}\] \[\Rightarrow 15{{r}^{2}}-14r+3=0\]\[\Rightarrow (3r-1)(5r-3)=0\]\[r=\frac{1}{3},\frac{3}{5}.\]        Only acceptable value is \[r=\frac{1}{3},\]because\[r\in \left( 0,\frac{1}{2} \right]\] \[\therefore \]\[c.d=7b-3a=7ar-3a=\frac{7}{3}a-3a=-\frac{2}{3}a\] \[\therefore \]\[{{4}^{th}}\,term=15c-\frac{2}{3}a=\frac{15}{9}a-\frac{2}{3}a=a\]


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