Answer: 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
Answer:
exponential decay
Step-by-step explanation:
Answer:
Step-by-step explanation:
\left[x _{2}\right] = \left[ \frac{-1+i \,\sqrt{3}+2\,by+\left( -2\,i \right) \,\sqrt{3}\,by}{2^{\frac{2}{3}}\,\sqrt[3]{\left( 432\,by+\sqrt{\left( -6912+41472\,by+103680\,by^{2}+55296\,by^{3}\right) }\right) }}+\frac{\frac{ - \sqrt[3]{\left( 432\,by+\sqrt{\left( -6912+41472\,by+103680\,by^{2}+55296\,by^{3}\right) }\right) }}{24}+\left( \frac{-1}{24}\,i \right) \,\sqrt{3}\,\sqrt[3]{\left( 432\,by+\sqrt{\left( -6912+41472\,by+103680\,by^{2}+55296\,by^{3}\right) }\right) }}{\sqrt[3]{2}}\right][x2]=⎣⎢⎢⎢⎢⎡2323√(432by+√(−6912+41472by+103680by2+55296by3))−1+i√3+2by+(−2i)√3by+3√224−3√(432by+√(−6912+41472by+103680by2+55296by3))+(24−1i)√33√(432by+√(−6912+41472by+103680by2+55296by3))⎦⎥⎥⎥⎥⎤
totally answer.
