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Tpy6a [65]
3 years ago
8

Express the number 220 as the sum of four numbers that form a geometric progression such that the third term is greater than the

first by 44.
Mathematics
2 answers:
damaskus [11]3 years ago
7 0
220 = x + xd + xd^2 + xd^3
The third term xd^2 is greater than the first term x by 44.
X+44=xd^2.
220= x+ xd + (x+44) + (x+44)d = 2x +2xd + 44d + 44

2x+2xd+44d=176
Divide both sides by 2
X+xd+22d=88
X+xd=88-22d
This is where I got stuck. Idk if I'm misinterpreting geometric progression.
siniylev [52]3 years ago
6 0

Answer:

Step-by-step explanation:

The first four terms of geometric series is:

a,ar,ar^2,ar^3

Since, we have given information that the third number is greater than 44 that means:

ar^2=a+44

Above equation can be rewritten as:

a(r^2-1)=44

Now, using:

a^2-b^2=(a+b)(a-b)

Here, a=r,b=1

a(r+1)(r-1)=44        (1)

The sum of first four terms is:

a+ar+ar^2+ar^3=220

a(1+r+r^2+r^3)=220

a(1(1+r)+r^2(1+r))=220

a((1+r)(1+r^2))=220      (2)

Divide equation (2) by (1) we get:

\frac{r^2+1}{r-1}=\frac{220}{44}

r^2+1=5r-5

\Rightarrow r^2-5r+6=0

\Rightarrow r^2-3r-2r+6=0

\Rightarrow r(r-3)-2(r-3)=0

\Rightarrow (r-2)(r-3)=0

\Rightarrow r=2,3

CASE1: When r=2 in ar^2=a+44

a(4)=a+44

3a=44

a=\frac{44}{3}

CASE2:When r=3 in ar^2=a+44

a(3)^2=a+44

\Rightarrow 9a=a+44

\Rightarrow 8a=44

\Rightarrow a=\frac{44}{8}=\frac{11}{2}

The series becomes:

From CASE1: \frac{44}{3},\frac{44\cdot 2}{3},\frac{44\cdot 2^2}{3},\frac{44\cdot 2^3}{3}

\Rightarrow \frac{44}{3},\frac{88}{3},\frac{176}{3},\frac{352}{3}

From CASE2: \frac{11}{2},\frac{11\cdot 3}{2},\frac{11\cdot 3^2}{2},\frac{11\cdot 3^3}{2}

\Rightarrow \frac{11}{2},\frac{33}{2},\frac{99}{2},\frac{297}{2}



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