Algebra

Let's solve this problem step-by-step.

1. Arithmetic Sequence and Sum Formulas

Let the arithmetic sequence be a1​,a2​,a3​,… with first term a1​ and common difference d.

The sum of the first n terms, Sn​, is given by:

Sn​=2n​[2a1​+(n−1)d]

2. Given Information

S20​=51​

S10​=0

3. Setting up Equations

Using the sum formula, we can write:

S20​=220​[2a1​+(20−1)d]=10(2a1​+19d)=51​

S10​=210​[2a1​+(10−1)d]=5(2a1​+9d)=0

4. Solving the Equations

From 5(2a1​+9d)=0, we have:

2a1​+9d=0

2a1​=−9d

a1​=−29​d

From 10(2a1​+19d)=51​, we have:

2a1​+19d=501​

Substitute 2a1​=−9d into 2a1​+19d=501​:

−9d+19d=501​

10d=501​

d=5001​

Now, substitute d=5001​ into 2a1​=−9d:

2a1​=−9(5001​)

2a1​=−5009​

a1​=−10009​

5. Find S_70

We want to find S70​:

S70​=270​[2a1​+(70−1)d]

S70​=35[2a1​+69d]

Substitute a1​=−10009​ and d=5001​:

S70​=35[2(−10009​)+69(5001​)]

S70​=35[−100018​+1000138​]

S70​=35[1000120​]

S70​=35[10012​]

S70​=35[253​]

S70​=25105​=521​

Therefore, S70​=521​.

s10  =   10a1  + (10 * 9)d / 2

s20   =   20a1  + (20 * 19)d / 2

So

0  =  10a1  + 45d    →  0  = -20a1 - 90d       (1)

1/5  = 20a1  +190d      (2)

add (1) , (2)

1/5  = 100d

d = 1/500

And

0  = 10a1 + 45(1/500)

10a1  =  -45/500

a1  = -45/5000  =  -9/1000

s70  =  70(-9/1000)  + (70 * 69)/2 * (1/500) =    -630/1000 +  2415 / 500  = 21 / 5