Question

7 The sum of the first $n$ terms of an arithmetic series is $A_n$, where $A_n = \sum_{r=1}^{n} (4r + 5)$ (a) For this arithmetic series, find (i) the first term, (ii) the common difference. The sum of the first $n$ terms of a geometric series is $G_n$, where $G_n = \sum_{r=1}^{n} 4(3)^{r-1}$ (b) For this geometric series, find (i) the first term, (ii) the common ratio. (c) Find the value of $n$ for which $A_{14} - 6 = G_n$

          7 The sum of the first $n$ terms of an arithmetic series is $A_n$, where
$A_n = \sum_{r=1}^{n} (4r + 5)$
(a) For this arithmetic series, find
(i) the first term,
(ii) the common difference.
The sum of the first $n$ terms of a geometric series is $G_n$, where
$G_n = \sum_{r=1}^{n} 4(3)^{r-1}$
(b) For this geometric series, find
(i) the first term,
(ii) the common ratio.
(c) Find the value of $n$ for which $A_{14} - 6 = G_n$

7 The sum of the first n terms of an arithmetic series is An, where
An = ∑r=1^n (4r + 5)
(a) For this arithmetic series, find
(i) the first term,
(ii) the common difference.
The sum of the first n terms of a geometric series is Gn, where
Gn = ∑r=1^n 4(3)^r-1
(b) For this geometric series, find
(i) the first term,
(ii) the common ratio.
(c) Find the value of n for which A14 - 6 = Gn

Added by Tina B.

Elementary and Intermediate Algebra

Alan S. Tussy, R. David Gustafson 5th Edition

Instant Answer

Solved by Expert Allison Knapp

Step 1

We are also given that 4 = Σ(4r+5), which means that the sum of the first n terms of the arithmetic series 4r+5 is 4. Using the formula for the sum of an arithmetic series, we can write: 4 = n/2 * (2(4r+5) + (n-1)d) Simplifying this equation, we get: 4 = n/2 Show more…

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The sum of the first terms of an arithmetic series is A, where 4 = Î£(4r+5) (a) For this arithmetic series, find the first term. the common difference The sum of the first terms of a geometric series is G, where 2 = Î£(4(3)^n) For this geometric series, find the first term, the common ratio (c) Find the value of n for which Ap = G