General formula for n-th term of arithmetical progression is
a(n)=a(1)+d(n-1).
For 3d term we have
a(3)=a(1) +d(3-1), where a(3)=7
7=a(1)+2d
For 7th term we have
a(7)=a(1) +d(7-1)
a(7)=a(1) + 6d
Also, we have that the <span>seventh term is 2 more than 3 times the third term,
a(7)=3*a(3)+2= 3*7+2=21+2=23
So we have, </span>a(7)=a(1) + 6d and a(7)=23. We can write
23=a(1) + 6d.
Now we can write a system of equations
23=a(1) + 6d
<span> - (7=a(1)+2d)
</span>16 = 4d
d=4,
7=a(1)+2d
7=a(1)+2*4
a(1)=7-8=-1
a(1)= - 1
First term a(1)=-1, common difference d=4.
Sum of the 20 first terms is
S=20 * (a(1)+a(20))/2
a(1)=-1
a(n)=a(1)+d(n-1)
a(20) = -1+4(20-1)=-1+4*19=75
S=20 * (-1+75)/2=74*10=740
Sum of 20 first terms is 740.
Answer:
406
Step-by-step explanation:
add 776 by 1444 then using the sum subtract it by 1814 to get answer
Answer:
True
This is completely the way you can do it. So the equality will be true.
Answer:
Step-by-step explanation:
<h3>Given</h3>
<u>Inequality</u>: (x-1)(x+2)(2x-7)≤0
<h3>Solution: </h3>
<u>If we solve the corresponding equation (x-1)(x+2)(2x-7)²= 0, we get roots </u>
<u>We need to consider the following 4 intervals: </u>
- (−∞; −2), [−2; 1], (1; 3.5), (3.5; ∞)
<u>1st interval</u> (−∞; −2)
- The expression (x-1)(x+2)(2x-7)² is positive as two of the multiples are negative and one is always positive (square number), and therefore does not satisfy the inequality.
<u>2nd interval</u> [−2; 1]
- The expression is negative as only one of the multiples is negative, and therefore the interval (−1; 2) satisfies the inequality.
<u>3rd interval</u> (1; 3.5)
- The expression is positive as all the multiples are positive. Therefore, the interval (1; 3.5) also does not satisfy the inequality.
<u>4th interval</u>
- The expression is positive as above, and therefore also does not satisfy the inequality.
<u>So, the answer to the inequality is:
</u>
Answer:
1. -6a
2. -8a
3. -5a
4. -a
5. -3a
6. -6a -8c -4
Step-by-step explanation:
All you need to do is combine like terms
