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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

Posted by

Vakul

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