A geometric progression has a third term of 20 and a sum of infinity which is 3 times the first term. Find the first term

the first term is 45

Step-by-step explanation:

Step 1: The formular for the nth term is ar^n-1 and since we're solving for the third term ar^3-1 = ar² which is equal to 20

So ar²= 20 and we make this equation i

Step 2: the formular for the sum of infinity is Sn = a/(1-r) and according the question the sum of infinity is equals to 3 *A ( A means the first term)

Further simplification gives us

3a= a/(1-r)

We simplify Futher by multiplying both sides by (1-r)

3a.(1-r) = a/(1-r).(1-r)

3a(1-r)=a

= 3a-3ar=a

Make 3ar the subject of the formula

3ar= 2a and lets make this equation (ii)

Step 3: From the first equation since ar²=20, let's make a the subject of the formula by dividing both sides by r²

Ar²/r² = 20/r²

A=20/r²lets make this equation (iii)

Step 4: subtitle a as 20/r² into equation (ii)

Since 3ar=2a

Then 3.(20/r²).r = 2. 20/r²

= (60/r²) .r = 40/r²

60/r = 40/r2

Multiplying both sides by r² gives us

60r = 40

Divide both sides by 20 gives us

= 3r= 2

Divide both sides by 3 gives us

= r = 2/3

Step 5: substitute r for 2/3 in equation i

ar²= 20

a. (2/3)2 = 20

= a.4/9 = 20

= 4a = 180

= a = 180/4

a = 45

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step-by-step explanation: