Question

Find the correct sum of each geometric sequence.

Answer 1

A geometric sequence with first term "a" and common ratio "r" has "nth" term:

ar^(n-1)

And the sum of a geometric sequence with "n" terms, first term "a," and common ratio "r" has the sum "a(r^n - 1)/r - 1.

1.) 765

2.) 300

3.) 1441

4.) 244

5.) 2101

Answer 2

Answer:

1.765

2.301

3.1441

4.183

5.2101

Step-by-step explanation:

We are given that

1.$a_1=3, a_8=384,r=2$

We know that sum of nth term in G.P is given by

$S_n=\frac{a(r^n-1)}{r-1}$ when r > 1

$S_n=\frac{a(1-r^n)}{1-r}$ when r < 1

n=8, r=2 a=3

Therefore,$S_8=\frac{3((2)^8-1)}{2-1}$ because r > 1

$S_8=3\times (256-1)=765$

1. Sum of given G.P is 765

2.$a_1=343,a_n=-1,r=-\frac{1}{7}$

nth term of G.P is given by the formula

$a_n=ar^{n-1}$

Therefore , applying the formula

$-1=343\times (\frac{-1}{7}}^{n-1}$

$\frac{-1}{343}=(\frac{-1}{7})^{n-1}$

$(\frac{-1}{7})^3=(\frac{-1}{7})^{n-1}$

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

$S_4=\frac{343(1-(\frac{-1}{7})^4)}{1-\frac{-1}{7}}$ where r < 1

$S_4=\frac{343(1+\frac{1}{343})}{\frac{8}{7}}$

$S_4=343\times\frac{344}{343}\times\frac{7}{8}$

$S_4=301$

3.$a_1=625, n=5,r=\frac{3}{5} < 1$

Therefore, $S_5=\frac{625(1-(\frac{3}{5})^5)}{1-\frac{3}{5}}$

$S_5=625\times \frac{3125-243}{3125}\times \frac{5}{2}$

$S_5=625\times\frac{2882}{3125}\times\frac{5}{2}$

$S_5=1441$

4.$a_1=4,n=5,r=-3$

$S_5=\frac{4(1-(-3)^5}{1-(-3)}$ where r < 1

$S_5=\frac{3(1+243)}{1+3}$

$S_5=3\times 61=183$

5.$a_1=2402,n=5,r=\frac{-1}{7}$

$S_5=\frac{2401(1-(\frac{-1}{7})^5)}{1-\frac{-1}{7}}$ r < 1

$S_5=\frac{2401(1+\frac{1}{16807})}{\frac{7+1}{7}}$

$S_5=2401\times\frac{16808}{16807}\times\frac{7}{8}$

$S_5=2101$