Answer: the first term is 45
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