Answer:
1.765
2.301
3.1441
4.183
5.2101
Step-by-step explanation:
We are given that
1.[tex]a_1=3, a_8=384,r=2[/tex]
We know that sum of nth term in G.P is given by
[tex]S_n=\frac{a(r^n-1)}{r-1}[/tex] when r > 1
[tex] S_n=\frac{a(1-r^n)}{1-r}[/tex] when r < 1
n=8, r=2 a=3
Therefore,[tex]S_8=\frac{3((2)^8-1)}{2-1}[/tex] because r > 1
[tex]S_8=3\times (256-1)=765[/tex]
1. Sum of given G.P is 765
2.[tex]a_1=343,a_n=-1,r=-\frac{1}{7}[/tex]
nth term of G.P is given by the formula
[tex]a_n=ar^{n-1}[/tex]
Therefore , applying the formula
[tex]-1=343\times (\frac{-1}{7}}^{n-1}[/tex]
[tex]\frac{-1}{343}=(\frac{-1}{7})^{n-1}[/tex]
[tex](\frac{-1}{7})^3=(\frac{-1}{7})^{n-1}[/tex]
When base equal on both side then power should be equal
Then we get n-1=3
n=3+1=4
Applying the formula of sum of G.P
[tex]S_4=\frac{343(1-(\frac{-1}{7})^4)}{1-\frac{-1}{7}}[/tex] where r < 1
[tex] S_4=\frac{343(1+\frac{1}{343})}{\frac{8}{7}}[/tex]
[tex] S_4=343\times\frac{344}{343}\times\frac{7}{8}[/tex]
[tex]S_4=301[/tex]
3.[tex]a_1=625, n=5,r=\frac{3}{5} < 1[/tex]
Therefore, [tex] S_5=\frac{625(1-(\frac{3}{5})^5)}{1-\frac{3}{5}}[/tex]
[tex]S_5=625\times \frac{3125-243}{3125}\times \frac{5}{2}[/tex]
[tex]S_5=625\times\frac{2882}{3125}\times\frac{5}{2}[/tex]
[tex]S_5=1441[/tex]
4.[tex]a_1=4,n=5,r=-3[/tex]
[tex]S_5=\frac{4(1-(-3)^5}{1-(-3)}[/tex] where r < 1
[tex] S_5=\frac{3(1+243)}{1+3}[/tex]
[tex]S_5=3\times 61=183[/tex]
5.[tex]a_1=2402,n=5,r=\frac{-1}{7}[/tex]
[tex]S_5=\frac{2401(1-(\frac{-1}{7})^5)}{1-\frac{-1}{7}}[/tex] r < 1
[tex] S_5=\frac{2401(1+\frac{1}{16807})}{\frac{7+1}{7}}[/tex]
[tex] S_5=2401\times\frac{16808}{16807}\times\frac{7}{8}[/tex]
[tex]S_5=2101[/tex]
