Question

44. The 10th term of a GP whose 8th term is 192 and the common ratio is 3 is

A. 1782
B. 1738
C. 1728
D. 1278​

Answer 1

GIVEN: 8TH term is 192 i.e,

$a_{8}$=192

common ratio(r)=3

TO FIND: 10TH TERM i.e,

$a_{10}$=?

SOLUTION:

let a be the first term of the G.P

∴$a_{8}=ar^{8-1} =ar^{7}$

⇒$ar^{7}=192$

given r=3, so on substituting the value of r, we will get,

$a(3)^{7}=192$

$a(3)^{7}=(2)^{6} 3$

⇒$a=\frac{(2)^{6}3 }{(3)^{7} }=\frac{(2)^{6} }{(3)^{6} } =(2/3)^{6}$

∴$a_{10}=ar^{10-1}=(2/3)^{6}*(3)^{9} =\frac{(2)^{6}*(3)^{9} }{(3)^{6} } =(2)^{6}*(3)^{3}$

⇒64×27

=1728

hence, the required 10th term is 1728.

so,(c) 1728 is correct.