Let a,b and c be in G.P. with common ratio r , where a\neq 0 and

0<r\leq \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 :

  • Option 1)

    \frac{2}{3}a

  • Option 2)

    5 a

  • Option 3)

    \frac{7}{3}a

  • Option 4)

    a

 

Answers (1)

Since a,b,c are in G.P. with common ratio r

then 

b = ar , c=ar^{2}

Also 3a, 7b and 15c are in A.P.

2\cdot7b=3a+15c

14b=3a+15c

=> 14ar=3a+15ar^{2}

=> 14r=3+15r^{2}

=> 0=3+15r^{2}-14r

=> 0=15r^{2}-14r+3

=> 3r(5r-3)-1(5r-3)=0

=> (3r-1)(5r-3)=0

=> r=\frac{3}{5},\frac{1}{3}

So, terms are 3a,7.a.\frac{3}{5}, 15.a(\frac{3}{5})^{2}

                       or

                       3a,7.a.\frac{1}{3}, 15.a(\frac{1}{3})^{2}

=> 3a,\frac{21a}{5}, \frac{27a}{5}   or   3a,\frac{7a}{3}, \frac{5a}{3}

So, 4th term 

\frac{27a}{5}+\frac{6a}{5}=\frac{33a}{5}   or   \frac{5a}{3}-\frac{2a}{3}=a

So, option (4) is correct.

 


Option 1)

\frac{2}{3}a

Option 2)

5 a

Option 3)

\frac{7}{3}a

Option 4)

a

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