A g.p has 6 terms, if the 3rd and 4th terms are 28 and_56 respectively find :the first and the sum of the g.p
https://www.mathsisfun.com/algebra/sequences-sums-geometric.html
or brute force
1
2
3 28
4 28 *r = 56 so r = 2
so the series is
1 7
2 14
3 28
4 56
5 112
6 224
now add :)
I don't know the workings pls help me out
To find the first term and the sum of a geometric progression (g.p.), we need to gather information from the given problem.
Let's assume that the first term of the g.p. is represented by 'a', and the common ratio is represented by 'r'. Since the g.p. has 6 terms:
The third term = a * r^2 = 28 ------- (Equation 1)
The fourth term = a * r^3 = 56 ------- (Equation 2)
Now, let's solve these equations to find the values of 'a' and 'r'.
Dividing Equation 2 by Equation 1, we get:
(a * r^3) / (a * r^2) = 56 / 28
r = 2
Plugging the value of r into Equation 1, we can solve for 'a':
a * (2^2) = 28
4a = 28
a = 7
Therefore, the first term (a) of the g.p. is 7 and the common ratio (r) is 2.
To find the sum of the g.p., we can use the formula:
Sum of a g.p. = (a * (r^n - 1)) / (r - 1)
Here, 'n' represents the number of terms in the g.p. In this case, n = 6.
Substituting the values into the formula, we get:
Sum of the g.p. = (7 * (2^6 - 1)) / (2 - 1)
Sum of the g.p. = (7 * (64 - 1)) / 1
Sum of the g.p. = (7 * 63)
Sum of the g.p. = 441
Therefore, the first term of the g.p. is 7 and the sum of the g.p. is 441.