Let $a_1$, $a_2$, . . . , $a_{10}$ be an arithmetic sequence. If $a_1 + a_3 + a_5 + a_7 + a_9 = 17$ and $a_2 + a_4 + a_6 + a_8 +
a_{10} = 15$, what is the value of $a_1$?
GIVING BRAINLIEST>
1 answer:
So
a be first term and d be common difference
- a+a+2d+a+4d+a+6d+a+9d=17
- 5a+21d=17--(1)
And
- a+d+a+3d+a+5d+a+7d+a+9d=15
- 5a+25d=15--(2)
Eq(1)-Eq(2)
Put in second one
- 5a+25d=15
- a+5d=3
- a+5/2=15
- a=15-5/2
- a=25/2
You might be interested in
Answer:
4.76
Step-by-step explanation:
constant of proportionality=slope
slope=m
y=mx+b
m=4.76
The answer to this question is 8.2 because absolute value is how far away a number is from zero.
Answer:
x= 5 , y = 34
Step-by-step explanation:
8x-6=5x+9
3x-6=9
3x=15
x=5
-
8(x5) - 6 =
34
25(x5) +9 =
34
I believe the answer is D
Answer:
jack is correct because is greater than 5
Step-by-step explanation: