Let the first term, common difference and number of terms of an AP are a, d and n respectively.
Given that, 9th term of an AP, T9 = 0 [∵ nth term of an AP, Tn = a + (n-1)d]
⇒ a + (9-1)d = 0
⇒ a + 8d = 0 ⇒ a = -8d ...(i)
Now, its 19th term , T19 = a + (19-1)d
= - 8d + 18d [from Eq.(i)]
= 10d ...(ii)
and its 29th term, T29 = a+(29-1)d
= -8d + 28d [from Eq.(i)]
= 20d = 2 × T19
Hence, its 29th term is twice its 19th term
Answer:
54
Step-by-step explanation:
$1284 + $910
= $2184
$2184 ÷ $40
= 54.6
= 54
Answer:
below
Step-by-step explanation:
n = 30
cos 2*30 = cos 60 = 1/2
sin 30 = 1/2
2m = 30
m=15
Answer:
x² - 16
Step-by-step explanation:
Given
(x - 4)(x + 4)
Each term in the second factor is multiplied by each term in the first factor, that is
x(x + 4) - 4(x + 4) ← distribute both parenthesis
= x² + 4x - 4x - 16 ← collect like terms
= x² - 16
I think it is a because it shows us the points that the students made. I also think it is d. But I think it is mainly a.
-Dhruva;)
