Question

Find the sum of the series:
10 + 14 + 18 + 22 +...+138
and 6-12+24-48+96-...+1536

Answer 1

The first one is an arithmetic series where a=10 and d=4
Use 138 as the last term to find the number of terms in the series, then use the sum of terms formula.

The second is a geometric series with a=6 and r=-2
Use 1536 to find the number of terms you have, then the sum of terms of a GS formula

Surely since you are studying this topic you must have the formulas required to answer the above.
Let me know what you got.

Answer 2

10 + 14 + 18 + 22 +...+138

arithmetic series
difference = d = 4
sum = (n/2)(a1 + an)
where
138 = 10 + 4(n-1)
128 = 4 n - 4
132 = 4 n
n = 33
so
sum = (33/2)(148)
=2442

geometric
6-12+24-48+96-...+1536

6 + (-2)6 + (-2)^2 (6) + (-2)^3 (6)
g = 6
r = -2

last term is g r^(n-1)
1536 = 6 (-2)^n-1)
256 = (-2)^(n-1)
log 256 = (n-1) log (-2)
well, lets hope n-1 is even so -2^n = 2^n. it is +1536 so it is OK
2.40824 = (n-1) .30103
n - 1 = 8
n = 9
so sum = 6(1- (-2)^9 )/ (1-(-2))
= 6(1+512)/3
=1026


An+1 = -2 An

Answer 3

ok for the first one i got stuck at:
n^2 + 4n - 69=0
now factoring doesn't work. what should i do??

second one
i get stuck at -4608=6(-2^n - 1)