3 years ago

Algebra 1 plz help me

Mathematics

1 answer:

sasho [114]3 years ago

8 0

Answer:

Step-by-step explanation:

solution is (4,3)

x=4

y=3

this point lies in both the tables.

we have to find a pair which lies on both.

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Find the sumofthe geometrical progression of five terms, of which the first term is 7 and the multiplier is 7.Verify that the su

777dan777 [17]

Answer:

The sum of first five term of GP is 19607.

Step-by-step explanation:

We are given the following in the question:

A geometric progression with 7 as the first term and 7 as the common ration.

$a, ar, ar^2,...\\a = 7\\r = 7$

$7, 7^2, 7^3, 7^4...$

Sum of n terms in a geometric progression:

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

For sum of five terms, we put n= 5, a = 7, r = 7

$S_5 = \displaystyle\frac{7(7^5 - 1)}{(7-1)}\\\\S_5 = 19607$

The sum of first five term of GP is 19607.

Verification:

$2801\times 7 = 19607$

Thus, the sum is equal to product of 2801 and 7.

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3 years ago