Question

Given a GP with a=729 and 7th term =64 find the 10 terms​

Answer 1

Answer:

$\sf{The \ first \ ten \ terms \ of \ the \ G.P. \ are}$

$\sf{729, \ 486, \ 324, \ 216, \ 144, \ 96, \ 64, \ 42.67,}$

$\sf{28.44, \ and \ 16.29 \ respectively. }$

Given:

$\sf{In \ a \ G.P.}$

$\sf{a=729 \ and \ a_{7}=64}$

To find:

$\sf{Ten \ terms \ of \ the \ G.P.}$

Solution:

$\boxed{\sf{a_{n}=ar^{n-1}}}$

$\sf{\therefore{a_{7}=ar^{7-1}}}$

$\sf{\therefore{64=729\times \ r^{6}}}$

$\sf{\therefore{r^{6}=\dfrac{64}{729}}}$

$\sf{\therefore{r^{6}=(\dfrac{2}{3})^{6}}}$

$\sf{\therefore{r=\dfrac{2}{3}}}$

$\sf{a_{1}=729,}$

$\sf{a_{2}=ar=729\times\dfrac{2}{3}=486,}$

$\sf{a_{3}=ar^{2}=729\times(\dfrac{2}{3})^{2}=324,}$

$\sf{a_{4}=ar^{3}=729\times(\dfrac{2}{3})^{3}=216,}$

$\sf{a_{5}=ar^{4}=729\times(\dfrac{2}{3})^{4}=144,}$

$\sf{a_{6}=ar^{5}=729\times(\dfrac{2}{3})^{5}=96,}$

$\sf{a_{7}=ar^{6}=729\times(\dfrac{2}{3})^{6}=64,}$

$\sf{a_{8}=ar^{7}=729\times(\dfrac{2}{3})^{7}=42.67,}$

$\sf{a_{9}=ar^{8}=729\times(\dfrac{2}{3})^{8}=28.44,}$

$\sf{a_{10}=ar^{9}=729\times(\dfrac{2}{3})^{9}=16.29}$

$\sf\purple{\tt{\therefore{The \ first \ ten \ terms \ of \ the \ G.P. \ are}}}$

$\sf\purple{\tt{729, \ 486, \ 324, \ 216, \ 144, \ 96, \ 64, \ 42.67,}}$

$\sf\purple{\tt{28.44, \ and \ 16.29 \ respectively. }}$