I am stuck on this question:
The sum of the first and the third terms of a Geometric Progression is $20$ and the sum of its first three terms is 26. Find the progression by writing the first $5$ terms.
I know these formula: $$S_n= \frac{a(1-r^n)}{(1-r)}$$
where the $n$-th term is $ar^{n-1}$.
However I don't know how to find the first term $(a)$, the ratio $(r)$.
any help will be very helpful.
Thank you.
$$a+ar^2=20,\\a+ar+ar^2=26.$$
Obviously, $$ar=6$$
leading to
$$20a=a(a+ar^2)=a^2+6^2.$$
The solutions of this quadratic equation are
$$a=2,r=3,\\a=18,r=\frac13.$$
Then
$$2,6,18,54,162,\cdots$$ or $$18,6,2,\frac23,\frac29,\cdots.$$
from the Problem we have $$a_1+a_3=20$$ and $$a_1+a_2+a_3=26$$ with $$a_2=a_1q$$ and $$a_3=a_1q^2$$ we get the System $$a_1+a_1q^2=20$$ and $$a_1+a_1q+a_1q^2=26$$ can you solve this? dividing both equations and assuming that $$a_1\ne 0$$ we get $$\frac{1+q^2}{1+q+q^2}=\frac{10}{13}$$ from here we get $$1+q^2=\frac{10}{13}(1+q+q^2)$$ can you solve this equation?
Let $a $ be the first term of the progression and $r $ be the common ratio. It is given that $$a + ar^2 =20\tag {1}$$ $$a+ar+ar^2 =26\tag {2}$$ $$(2)-(1) \implies ar =6 \tag {3}$$ Thus, substituting in $(1) $, we get, $$\frac {6}{r}+6r =20 \Rightarrow 6r^2-20r+6=0$$ $$\Rightarrow (6r-2)(r-3)=0$$ $$\Rightarrow r =\frac {1}{3},3$$
Hope you can take it from here by solving for $a $ in $(3) $ and writing out the terms.
continuing (I will write $a$ instead of $a_1$):
since $a+aq^2=20$ and $a+aq+aq^2=26$, you have $aq+20=26$, i.e. $aq=6$.
now $20=a+aq^2=a+aq\cdot q=a+6q$.
the system of equations $a+6q=20$ and $aq=6$ leads to $a+\frac{36}{a}=20$, what gives $a=2$ or $a=18$. Hence $q=3$ or $q=\frac{1}{3}$ respectively.