Arithmatic and Geometric Progression Please Help

[s021s]

1. The sum of the first n temrs of a certin arithmetic progression is given by Sn ( big s and little n under it ) = N2 ( n to the power of 2 ) + 3n

a) Find Expression for Sn-1 ( big s little n underit with - 1) and deduce an expression for Un ( big u with little n under it )
b) find the first term and the common difference

2. the lengths of the sides of a triangle are in geometric progression. the length of the shortest side is 6cm and the primeter of the triangle is 28 1/2 ( 28 and 0.5 ) find the lenth of the other sides.

3. in a geometric progression the sum to infinity 9 and the sum of the first two terms is 5. find the first 4 terms of the porogression given that they are positive .

 

[soroban]

Hello, s021s!

2. The lengths of the sides of a triangle are in geometric progression.
The length of the shortest side is 6 cm and the perimeter of the triangle is 28.5 cm.
Find the length of the other sides.

\(\displaystyle \text{The first 3 terms of a geometric progression are: }\:a,\:ar,\:ar^2\)

\(\displaystyle \text{We are told that }a = 6.\)

\(\displaystyle \text{The three terms are: }\;6,\:6r,\;6r^2\)

\(\displaystyle \text{The perimeter is 28.5: }\;6 + 6r + 6r^2 \:=\:\tfrac{57}{2} \quad\Rightarrow\quad 4r^2 + 4r - 15 \:=\:0\)

. . \(\displaystyle (2r-3)(2r+5) \:=\:0 \quad\Rightarrow\quad r = \tfrac{3}{2},\;\rlap{////}-\tfrac{5}{2}\)

\(\displaystyle \text{The three sides are: }\;\begin{Bmatrix}a &=& 6 \\ ar &=& 9 \\ ar^2 &=& \frac{27}{2} \end{Bmatrix}\)

3. in a geometric progression, the sum to infinity is 9, and the sum of the first two terms is 5.
Find the first 4 terms of the progression given that they are positive .

\(\displaystyle \text{The sum to infinity is 9: }\;\frac{a}{1-r} \:=\:9 \quad\Rightarrow\quad a \:=\:9(1-r)\;\;[1]\)

\(\displaystyle \text{Sum of first two terms is 5: }\;a + ar \:=\:5 \quad\Rightarrow\quad a(1+r) \:=\:5 \quad\Rightarrow\quad a \:=\:\frac{5}{1+r}\;\;[2]\)

\(\displaystyle \text{Equate [2] and [1]: }\;\frac{5}{1+r} \:=\:9(1-r) \quad\Rightarrow\quad 5 \:=\:9(1-r)(1+r)\)

. . . \(\displaystyle \tfrac{5}{9} \:=\:1-r^2 \quad\Rightarrow\quad r^2 \:=\:\tfrac{4}{9} \quad\Rightarrow\quad r \:=\:\pm\tfrac{2}{3}\)

\(\displaystyle \text{Since the terms are all positive, }\:r \:=\:\tfrac{2}{3}\)

\(\displaystyle \text{Substitute into [1]: }\;a \:=\:9(1 - \tfrac{2}{3}) \quad\Rightarrow\quad a\:=\:3\)


\(\displaystyle \text{The first four terms are: }\;3,\:2,\:\tfrac{4}{3},\:\tfrac{8}{9}\)

 

[soroban]

Hello again, s021s!

I don't think there is enough information to solve #1 . . .

\(\displaystyle \text{1. The sum of the first }n\text{ terms of a certain arithmetic progression is given by: }\;S_n \;=\;n^2 \,+\, 3n\)

\(\displaystyle \text{a) Find an expression for }S_{n-1}\)

\(\displaystyle S_{n-1} \;\;=\;\;(n-1)^2 + 3(n-1) \;\;=\;\; n^2 - 2n + 1 + 3n - 3\)

\(\displaystyle \boxed{S_{n-1}\;\;=\;\;n^2 + n - 2}\)

\(\displaystyle \text{(b) Find the first term and the common difference.}\)

\(\displaystyle \text{The common difference is: }\;d \;=\;S_n - S_{n-1}\)

. . \(\displaystyle d \;=\;(n^2+3n) - (n^2+n-2) \quad\Rightarrow\quad \boxed{d \;=\;2(n+1)}\)


\(\displaystyle \text{The sum of the first }n\text{ terms is: }\;S_n \;=\; \frac{n}{2}\bigg[2a + (n-1)d\bigg]\)

\(\displaystyle \text{So we have: }\;\frac{n}{2}\bigg[2a + (n-1)2(n+1)\bigg] \;=\;n^2 + 3n \quad\Rightarrow\quad n[a + n^2-1] \;=\;n^2 + 3n\)

. . . . . . . . . \(\displaystyle na + n^3 - n \;=\;n^2 + 3n \quad\Rightarrow\quad na \;=\;4n + n^2 - n^3\)

\(\displaystyle \text{The first term is: }\;\boxed{a \;=\;4 + n - n^2}\)

\(\displaystyle \text{(c) Deduce an expression for }U_n\)

\(\displaystyle U_n \;=\;a + (n-1)d\)

\(\displaystyle U_n \;=\;(4+n-n^2) + (n-1)2(n+1) \;=\;4 + n - n^2 + 2n^2 - 2\)

\(\displaystyle \boxed{U_n \;=\;n^2 + n + 2}\)


~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


\(\displaystyle \text{And that's the best I can do.}\)

\(\displaystyle \text{All answers are in terms of }n.\)

 

[Deleted member 4993]

S[sub:2xjz0ydo]1[/sub:2xjz0ydo] = first term = 4

S[sub:2xjz0ydo]2[/sub:2xjz0ydo] - S[sub:2xjz0ydo]1[/sub:2xjz0ydo] = second term = 6

S[sub:2xjz0ydo]3[/sub:2xjz0ydo] - S[sub:2xjz0ydo]2[/sub:2xjz0ydo] = third term = 8

S[sub:2xjz0ydo]n-1[/sub:2xjz0ydo] - S[sub:2xjz0ydo]n-2[/sub:2xjz0ydo] = a[sub:2xjz0ydo]n-1[/sub:2xjz0ydo] =(n-1)[sup:2xjz0ydo]2[/sup:2xjz0ydo] + 3(n-1) - (n-2)[sup:2xjz0ydo]2[/sup:2xjz0ydo] - 3(n-2) = 2n

S[sub:2xjz0ydo]n[/sub:2xjz0ydo] - S[sub:2xjz0ydo]n-1[/sub:2xjz0ydo] = a[sub:2xjz0ydo]n[/sub:2xjz0ydo] =2n + 2

a[sub:2xjz0ydo]1[/sub:2xjz0ydo] = 4 and d = 2 [? a[sub:2xjz0ydo]n[/sub:2xjz0ydo] - a[sub:2xjz0ydo]n-1[/sub:2xjz0ydo]]