A plausible guess might be that the sequence is formed by a degree-4* polynomial,
From the given known values of the sequence, we have
Solving the system yields coefficients
so that the n-th term in the sequence might be
Then the next few terms in the sequence could very well be
It would be much easier to confirm this had the given sequence provided just one more term...
* Why degree-4? This rests on the assumption that the higher-order forward differences of 
eventually form a constant sequence. But we only have enough information to find one term in the sequence of 4th-order differences. Denote the k-th-order forward differences of by 
. Then
• 1st-order differences:
• 2nd-order differences:
• 3rd-order differences:
• 4th-order differences:
From here I made the assumption that 
is the constant sequence {15, 15, 15, …}. This implies 
forms an arithmetic/linear sequence, which implies 
forms a quadratic sequence, and so on up forming a quartic sequence. Then we can use the method of undetermined coefficients to find it.
Answer:
Sum of the first 15 terms = -405
Step-by-step explanation:
a + 3d = -15 (1)
a + 8d = -30 (2)
Where,
a = first term
d = common difference
n = number of terms
Subtract (1) from (1)
8d - 3d = -30 - (-15)
5d = -30 + 15
5d = -15
d = -15/5
= -3
d = -3
Substitute d = -3 into (1)
a + 3d = -15
a + 3(-3) = -15
a - 9 = -15
a = -15 + 9
a = -6
Sum of the first 15 terms
S = n/2[2a + (n − 1) × d]
= 15/2 {2×-6 + (15-1)-3}
= 7.5{-12 + (14)-3}
= 7.5{ -12 - 42}
= 7.5{-54}
= -405
Sum of the first 15 terms = -405
Answer:
Step-by-step explanation:
-5+3 + (-1/6)+ 5/6
-2 - 1/6 + 5/6
-2 + 4/6 (do this boring math)
300km because if 1cm=50km And it’s 6cm you have to do 50*6 to get C. 300km
