Answer:
Step-by-step explanation:
t1 = a1
t2 = a1 + d
t3 = a1 + 2d
The sum is t1 + t2 + t3 = a1 + a1 + d + a1 + 2d = 3a1 + 3d
Second question
t1 = a1
t2 = a1 + d
t3 = a1 + 2d
t4 = a1 + 3d
...
tn = a1 + (n - 1)*d
Sum = (a1 + a1 + (n- 1)*d) * n / 2
Sum = (2a1 + nd - d) * n / 2
Sum = (2a1*n + n^2*d - nd) / 2
Answer:
The last term is 33.
Step-by-step explanation:
Sn = (n/2)(a1 + L) where a1 = first term and L = the last.
So:
-480 = (20/2) ( -81 + L)
-480 = 10( L - 81)
L- 81 = -480 / 10 = -48
L = -48 + 81
L = 33.
(5300 + 1335.6) = 5300(1 + 0.07)^n where n is the number of years required
6635.6 = 5300(1.07)^n
(1.07)^n = 1.252
n log 1.07 = log 1.252
n = log 1.252 / log 1.07 = 3.32 years Answer
60% please make me as the most brainlyest
BC² = 11² + 11²
BC² = 121 + 121
BC² = 242
BC = √242
BC = 11√2
Answer: 11√2
