Answer:
<h2>
The eleventh term of the sequence is 64</h2>
Step-by-step explanation:
The sequence given is an arithmetic sequence
14, 19, 24, …………., 264
The nth term of an arithmetic sequence is given as;
Tn = a+(n-1)d where;
a is the first term = 14
d is the common difference = 19-14=24-19 = 5
n is the number of terms = 11(since we are to look for the eleventh term of the sequence)
substituting the given values in the formula given;
T11 = 14+(11-1)*5
T11 = 14+10(5)
T11 = 14+50
T11 = 64
The eleventh term of the sequence is 64
Can you add the picture to the question?
X/2-y/3=3/2
(6×x/2)-(6×y/3)=6×3/2
3x-2y=9______(1)
x/3+y/2=16/3
(6×x/3)+(6×y/2)=6×16/3
2x+3y=32_____(2)
(1)×3____9x-6y=27____(3)
(2)×2____4x+6y=64____(4)
(3)+(4)___13x=91
x=7
3(7)-2y=9
-2y=-12
y=6
Answer:
D
Step-by-step explanation:
The recursive formula allows us to find any term in a sequence by adding the common difference d to the previous term.
Here d = - 1 - (- 7) = - 1 + 7 = 6 , then
= 
+ 6 ; a₁ = - 7
