Question

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.

Answer 1

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.

Answer 2

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: $\frac{11}{2},\frac{11\cdot 3}{2},\frac{11\cdot 3^2}{2},\frac{11\cdot 3^3}{2}$

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