# How do you find the first five terms in the geometric sequence which is such that the sum of the 1st and 3rd terms is 50, and the sum of the 2nd and 4th terms is 150?

Nov 8, 2015

$5 , 15 , 45 , 135 , 405$

#### Explanation:

${A}_{n} = {A}_{1} {r}^{n - 1}$

${A}_{1} + {A}_{3} = 50$

$\implies {A}_{1} + {A}_{1} {r}^{3 - 1} = 50$

$\implies {A}_{1} + {A}_{1} {r}^{2} = 50$

${A}_{2} + {A}_{4} = 150$

$\implies {A}_{1} {r}^{2 - 1} + {A}_{1} {r}^{4 - 1} = 150$

$\implies {A}_{1} r + {A}_{1} {r}^{3} = 150$

$\implies r \left({A}_{1} + {A}_{1} {r}^{2}\right) = 150$

But ${A}_{1} + {A}_{1} {r}^{2} = 50$

$\implies r \left(50\right) = 150$

$\implies r = 3$

Now we have the common ratio. Next, we get the first term

${A}_{1} + {A}_{3} = 50$

$\implies {A}_{1} + {A}_{1} {r}^{2} = 50$

$\implies {A}_{1} + {A}_{1} \left({3}^{2}\right) = 50$

$\implies {A}_{1} + 9 {A}_{1} = 50$

$\implies 10 {A}_{1} = 50$

$\implies {A}_{1} = 5$

Hence, our geometric sequence has the formula

${A}_{n} = 5 \cdot {3}^{n - 1}$

To get the ${n}^{t h}$ element, simply plugin its order into the formula

${A}_{1} = 5$
${A}_{2} = 5 \cdot {3}^{1} = 15$
${A}_{3} = 5 \cdot {3}^{2} = 45$
${A}_{4} = 5 \cdot {3}^{3} = 135$
${A}_{5} = 5 \cdot {3}^{4} = 405$