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Setler79 [48]
3 years ago
7

Find the correct sum of each geometric sequence.

Mathematics
2 answers:
Alex777 [14]3 years ago
5 0
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
grin007 [14]3 years ago
3 0

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

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