The first, third and ninth terms of an A.P are the first three terms of a G.P. If the 7th term of the A.P is 28, calculate the 20th term of the G.P and the sum of the first 12 terms of the G.P.?

Tifepraise

3 years ago

Part (a)

Let {a1,a2,a3,...} be the arithmetic progression (AP) and {b1,b2,b3,...} be the geometric progression (GP)

If a1 = x, then
a2 = x+d for some constant d
a3 = x+2d
a4 = x+3d
a5 = x+4d
a6 = x+5d
a7 = x+6d = 14, given 7th term of AP
a8 = x+7d
a9 = x+8d
We add on d each time to get the next term.

x+6d = 14 solves to x = 14-6d
We'll use this equation later. Let's call this equation (1).

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Since the first, third and ninth terms of the AP are equal to the first three terms of the GP, this means
a1 = b1
a3 = b2
a9 = b3

The first three terms of the GP are
b1 = x
b2 = x*r
b3 = x*r^2
We multiply each term by r to get the next term.

Earlier we found that a3 = x+2d, which equates to b2 to get
a3 = b2
x+2d = x*r
14-6d+2d = (14-6d)*r ..... plug in x = 14-6d, which was equation (1).
14-4d = (14-6d)*r
r = (14-4d)/(14-6d) ... divide both sides by (14-6d)
We'll use this later. Call this equation (2).

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Also,
a9 = b3
x+8d = b3 ..... plug in a9 = x+8d
x+8d = x*r^2 .... plug in b3 = x*r^2
(14-6d)+8d = (14-6d)*r^2 ... plug in equation (1)
14+2d = (14-6d)*r^2
14+2d = (14-6d)*((14-4d)/(14-6d))^2 .... plug in equation (2).
14+2d = ((14-4d)^2)/(14-6d) .... one pair of (14-6d) terms cancel
(14+2d)(14-6d) = (14-4d)^2 .... multiply both sides by (14-6d)
196-84d+28d-12d^2 = 196-56d-56d+16d^2 .... use the FOIL rule to expand
196-56d-12d^2 = 196-112d+16d^2 ... combine like terms
0 = 196-112d+16d^2-196+56d+12d^2 .... get everything to one side
0 = 28d^2-56d ... combine like terms
28d^2-56d = 0
28d(d-2) = 0 ... factor
28d = 0 or d-2 = 0 .... zero product property
d = 0 or d = 2
If d = 0, then the AP is {14, 14, 14, 14, 14, 14, ...} which is trivial. Likely your teacher will want you to focus on the more interesting solution.


If d = 2, then
x = 14-6d
x = 14-6*2
x = 2
Therefore, a1 = 2 is the first term of the AP.

The nth term of the AP is
a(n) = a1 + d(n-1)
a(n) = 2 + 2(n-1)
a(n) = 2 + 2n-2
a(n) = 2n
A few items of the sequence are: {2, 4, 6, 8, 10, 12, 14, 16, 18, ...}
The first, third and ninth terms are highlighted in blue
Note that the sequence 2,6,18 shows up in part (b) as the first three terms of the GP.

Plug in n = 20
a(n) = 2n
a(20) = 2*20
a(20) = 40
The 20th term of the AP is 40

Part (b)

Use the value of d to calculate r
r = (14-4d)/(14-6d)
r = (14-4*2)/(14-6*2)
r = (14-8)/(14-12)
r = 6/2
r = 3

The first term of the GP is
b1 = x
b1 = 2

Calculate the sum of the first 12 terms of the GP. Starting term is a = 2 and common ratio is r = 3
Sn = a*(1-r^n)/(1-r)
S12 = 2*(1-3^12)/(1-3)
S12 = 2*(1-531441)/(1-3)
S12 = 2*(-531440)/(-2)
S12 = 531440

The long way to get this answer is to generate the first 12 terms
b1 = 2
b2 = 6 ... multiply previous term by r = 3 to get the next term
b3 = 18
b4 = 54
b5 = 162
b6 = 486
b7 = 1458
b8 = 4374
b9 = 13122
b10 = 39366
b11 = 118098
b12 = 354294
Then add up the terms
2+6+18+54+162+486+1458+4374+13122+39366+118098+354294 = 531440