Answer to Question #289689 in Algebra for Alvine

Question #289689

In a geometric progression, the sum of the second and third term is 6 and the third and fourth term is -12, find the first, the common ratio and the sum of the first 10 terms of the progression


1
Expert's answer
2022-01-25T06:50:50-0500

The general term of a GP is:


Tn=arn1T_n = ar^{n-1}

Therefore,


T2=ar21=arT3=ar31=ar2T4=ar41=ar3T_2 = ar^{2-1} = ar\\ T_3 = ar^{3-1} = ar^2\\ T_4 = ar^{4-1} = ar^3

Now,


T2+T3=6    ar+ar2=6(i)T3+T4=12    ar2+ar3=12(ii)T_2+T_3 = 6 \implies ar+ar^2 =6 \qquad \qquad(i)\\ T_3+T_4 = -12 \implies ar^2+ar^3 =-12 \quad \quad(ii)

From (ii):


r(ar+ar2)=12(iii)r(ar+ar^2) = -12 \qquad \qquad (iii)

put (i) in (iii):


6r=12    r=-26r =-12 \implies \textbf{r=-2}

Put r=-2 in (i):


a(2)+a(2)2=62a+4a=62a=6a=3a(-2) + a(-2)^2=6\\ -2a +4a=6\\ 2a=6\\ \textbf{a=3}

The sum of first n term of a GP is:


Sn=a1rn1r Given n=10, and putting other values to above formula: S10=31(2)101(2)S10=3(341)S10=1023S_{n}=a \cdot \frac{1-r^{n}}{1-r}\\ \text{ Given n=10, and putting other values to above formula: }\\ \begin{aligned} S_{10} &=3 \cdot \frac{1-(-2)^{10}}{1-(-2)} \\ S_{10} &=3 \cdot(-341) \\ S_{10} &=-1023 \end{aligned}


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