Mr. Zhang has two stamps of 50 and 80. He can pay with them______ How much do you need to pay to send a letter

Because the stamps with 50 points and 80 points each can be combined into six different postage rates: 50 + 50 = 100 (points), 80 + 80 = 160 (points), 50 + 80 = 130 (points), 50 + 50 + 80 = 180 (points), 50 + 80 + 80 = 210 (points), 50 + 50 + 80 + 80 = 260 (points), plus the two denominations of 50 points and 80 points, a total of 6 + 2 = 8 different postage rates can be paid

Mr. Zhang has two 50 cent stamps and two 80 cent stamps. How much postage can he pay with these stamps?
(sending a letter is the amount of money you have to pay.),

1. 50 points and 80 points 2. 2 sheets 50 points 3. 2 sheets 80 points 4. 2 sheets 50 points and 1 sheet 80 points 5. 2 sheets 80 points and 1 sheet 50 points 6. 50 points 7. 80 points 8. 2 sheets 50 points and 2 sheets 80 points

Given that the tolerance of the arithmetic sequence an is a positive integer, and A3 * A7 = - 12 A4 + A6 = - 4, then the sum of the first ten terms SN=

∵ an is an arithmetic sequence
∴a3+a7=a4+a6=-4 ①
And A3 × A7 = - 12
The solution of simultaneous ① and ②
A3 = - 6, a7 = 2 or A3 = 2, a7 = - 6
And the tolerance D ＞ 0,
∴a3=-6 ,a7=2
So 4D = a7-a3 = 8
D = 2, A1 = a3-2d = - 6-2 × 2 = - 10
an=a1+(n-1)d=-10+2(n-1)=2n-12
a10=2×10-12=8
S10=10(a1+a10)/2
=5(-10+8)
=-10
Answer: - 10

It is known that the arithmetic sequence {an} satisfies the following conditions: A3 = 7, A5 + A7 = 26, and the sum of the first n terms of {an} is SN

Let the tolerance of arithmetic sequence {an} be D, then A3 = a1 + 2D = 7a5 + A7 = 2A1 + 10d = 26, the solution is A1 = 3D = 2, ∧ an = 3 + 2 (n-1) = 2n + 1sn = n (3 + 2n + 1) 2 = N2 + 2n

It is known that the arithmetic sequence [an] satisfies the following conditions: A3 = 7, A5 + A7 = 26;
(2) BN = 1 / (an ^ 2-1) (n belongs to n *), find the first n terms and TN of sequence [BN]

a5=a3+2d,a7=a3+4d
a5+a7=2a3+6d
26=2*7+6d
d=2
a4=a3+d=7+2=9
a1=a3-2d=7-2*2=3
Sn=na1+n(n-1)d/2=n^2+2n
an=a1+nd=3+2n
bn=1/(an^2-1)=1/[4(n+2)(n+1)]=1/4[(1/(n+1)-1/(n+2)]

Let the sum of the first n terms of the arithmetic sequence {an} be Sn, a1 + A7 = 16, A3 + A4 = 14, and find SN

Isochromatic
3+4=1+6
So a1 + A6 = A3 + A4 = 14
a1+a7=16
subtract
a7-a6=2
That is, d = 2
a7=a1+6d=a1+12
So a1 + A7 = 2A1 + 12 = 16
a1=2
an=a1+(n-1)d=2n
So Sn = (a1 + an) n / 2 = n & sup2; + n

The known sequence {an} is an arithmetic sequence, the sum of the first n terms is Sn, A3 = 7, S4 = 24
Finding the general term formula of sequence {an}

a3 = a1 + 2d = 7
S4 = (a4 + a1)*4/2 = (a3 +d +a1)*2 = (7 +a1 +d)*2 = 24
a1 + d = 5
d = 2
a1 = 3
an = 3 + 2(n-1) = 2n+1

Arithmetic sequence d = 2, n = 15, an = - 10 for A1 and Sn

an=a1+（n-1）*d
Take n = 15, d = 2, an = - 10 in
There are
a1+（15-1）*2=-10
The solution is: A1 = - 38
Sn = n * a1 + n * (n-1) * D / 2 or Sn = n * (a1 + an) / 2
Then n = 15, A1 = - 38, d = 2, an = - 10 are substituted into the above formula
Get Sn = - 360

In the arithmetic sequence an, given D = 1, an = 3, Sn = - 15, find A1 and n

Sn=(a1+an)n/2
So - 15 = (a1 + 3) n / 2
an=a1+d(n-1)=a1+n-1=3
So A1 = 4-N
So (4-N + 3) n / 2 = - 15
n²-7n-30=0
(n-10)(n+3)=0
therefore
n=10
a1=4-n=-6

We know that in the arithmetic sequence [an], d = 2, an = 1, Sn = - 15, find n and A1

If the sequence is arranged in reverse order, then an = 1 is the first term, Sn = - 15 remains unchanged, the number of terms n remains unchanged, and the tolerance becomes - D = - 2
Using the formula, we get the following results
n×1+n(n-1)×(-2)/2=-15
n^2-2n-15=0
N = 5, n = - 3 (rounding off)
1=a1+(5-1)×2
a1=-7

Mr. Zhang has two stamps of 50 and 80. He can pay with them______ How much do you need to pay to send a letter