Question

9. The fourth term of a G. P. is 27 and the 7th term is 729, find the gp​

Answer 1

Answer:

$\sf{The \ required \ G.P. \ is \ 1, \ 3, \ 9,...}$

Given:

$\sf{\leadsto{The \ forth \ term \ of \ a \ G.P. \ is \ 27}} \sf{and \ the \ 7^{th} \ term \ is \ 729}$

To find:

$\sf{\longmapsto{The \ G.P.}}$

Solution:

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

$\sf{ According \ to \ the \ first \ condition} \\ \\ \\ \sf{a_{4}=ar^{4-1}} \\ \\ \sf{\therefore{27=ar^{3}...(1)}} \\ \\ \sf{According \ to \ the \ second \ condition} \\ \\ \\ \sf{a_{7}=ar^{7-1}} \\ \\ \sf{\therefore{729=ar^{6}}} \\ \\ \sf{From \ (1) \ we \ get \ a=\dfrac{27}{r^{3}}} \\ \\ \\ \sf{Substitute \ a=\dfrac{27}{r^{3}} \ in \ equation (2), \ we \ get} \\ \\ \sf{729=(\dfrac{27}{r^{3}})\times \ r^{6}} \\ \\ \\ \sf{\therefore{r^{3}=\dfrac{729}{27}}} \\ \\ \sf{\therefore{r^{3}=27}} \\ \\ \sf{\therefore{r=3}} \\ \\ \sf{Substitute \ r=3 \ in \ equation (1), \ we \ get} \\ \\ \sf{27=a\times3^{3}} \\ \\ \sf{\therefore{a=\dfrac{27}{27}}} \\ \\ \sf{\therefore{a=1}}} \\ \\ \sf{\therefore{a_{1}=1}} \\ \\ \sf{a_{2}=ar=3} \\ \\ \sf{a_{3}=ar^{2}=9} \\ \\ \\ \purple{\tt{\therefore{The \ required \ G.P. \ is \ 1, \ 3, \ 9,...}}}$