Answer:
15th term = 116
Step-by-step explanation:
a= 4
Sum of an A.P = n/2 {2a + (n-1)d}
sum of first five term is equal to one-fourth of the sum of the next five term
5/2{ 2*4 + (5-1)d} = 1/4 × 10/2{2*4 + (10-1)d
5/2 {8 + 4d} = 1/4 × 5{ 8 + 9d}
40/2 + 20/2d = 1/4{ 40 + 45d)
20 + 10d= 40/4 + 45/4d
20 + 10d = 10 + 45/4d
20 - 10 = 45/4d - 10d
10 =45d - 40d /4
10 = 5/4d
Divide both sides by 5/4
10 ÷5/4 = d
10×4/5 = d
40/5 = d
8 = d
d= 8
Find the 15th term
15th term = a + (n-1)d
= 4 + (15-1)8
= 4 + (14)8
= 4 + 112
= 116
The 15th term is 116
Answer:
1.27
Step-by-step explanation:
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There is an infinite amount of 10 digits numbers with at leas two 1s in it.
Answer:
Step-by-step explanation:
5x + 9y = -11
3x + 9y = -3
5x + 9y = -11
-3x - 9y = 3
2x = -8
x = -4
-12 + 9y = -3
9y = 9
y = 1
(-4, 1)
Answer is C
Answer:It doesn't make sense when I try to do it,I even tried a calculator
Step-by-step explanation: IDK?