The sum of the first and third term is 40, while the addition of the second and fourth term is 96. How do you find the sixth term of the sequence?

Nov 8, 2015

Assuming the sequence is arithmetic, ${A}_{6} = 132$.

Explanation:

NOTE: This answer assumes that the sequence is arithmetic.
If the sequence is geometric, the answer is not a very nice number. Hence, I concluded the question must be dealing with an arithmetic sequence. If this is not the case, please inform me and i'll update the answer.

${A}_{n} = {A}_{1} + d \left(n - 1\right)$

${A}_{1} + {A}_{3} = 40$

$\implies {A}_{1} + \left({A}_{1} + d \left(3 - 1\right)\right) = 40$
$\implies {A}_{1} + {A}_{1} + 2 d = 40$
$\implies 2 {A}_{1} + 2 d = 40$

${A}_{2} + {A}_{4} = 96$

$\implies \left({A}_{1} + d \left(2 - 1\right)\right) + \left({A}_{1} + d \left(4 - 1\right)\right) = 96$

$\implies {A}_{1} + d + {A}_{1} + 3 d = 96$

$\implies 2 {A}_{1} + 4 d = 96$

$\implies 2 {A}_{1} + 2 d + 2 d = 96$

$\implies 40 + 2 d = 96$

$\implies 2 d = 56$
$\implies d = 28$

Now that we know the common difference, we need to find the first term

${A}_{1} + {A}_{3} = 40$

$\implies 2 {A}_{1} + 2 d = 40$

$\implies 2 {A}_{1} + 2 \left(28\right) = 40$

$\implies 2 {A}_{1} + 56 = 40$
$\implies 2 {A}_{1} = - 16$

$\implies {A}_{1} = - 8$

To find the sixth term, we use the general formula...

${A}_{n} = {A}_{1} + d \left(n - 1\right)$

$\implies {A}_{6} = {A}_{1} + d \left(6 - 1\right)$
$\implies {A}_{6} = - 8 + 28 \left(5\right)$
$\implies {A}_{6} = - 8 + 140$
$\implies {A}_{6} = 132$