The sum of 1st five terms of an AP nd the sum of the 1st seven termd of the same AP is 167. If the sum of the 1st ten terms of this AP is 235, find the sum of its1st twenty terms

5a + 10d + 7a + 21d = 167

10a + 45d = 235

solve the system for a and d

Sā‚‚ā‚€ = 20a + 190d

To find the sum of the first 20 terms of an arithmetic progression (AP), we need to use the formula for the sum of the first n terms of an AP:

S_n = (n/2) * (2a + (n-1)d)

Where:
S_n is the sum of the first n terms
n is the number of terms
a is the first term
d is the common difference

Let's break down the given information step by step:

1. We have the sum of the first five terms and seven terms of the same AP as 167 and know that:
S_5 = 167
S_7 = 167

2. We also know that the sum of the first ten terms is 235:
S_10 = 235

Now, we can solve for the values of a and d using the given information.

Step 1: Solving for a
We can find a by substituting the values of S_5 and S_7 into the sum formula and solving a system of equations.

S_5 = (5/2)(2a + (5-1)d) = 167
Simplifying: 5a + 10d = 334

S_7 = (7/2)(2a + (7-1)d) = 167
Simplifying: 7a + 21d = 334

Now, we have a system of equations:
5a + 10d = 334
7a + 21d = 334

Multiplying the first equation by 7 and the second equation by 5 to eliminate a:
35a + 70d = 2338
35a + 105d = 1670

Subtracting the equations:
35a + 105d - 35a - 70d = 1670 - 2338
35d = -668
d = -668/35
d = -19.09 (approx.)

Step 2: Solving for a (continued)
Substituting the value of d back into either of the original equations:
5a + 10(-19.09) = 334
5a - 190.9 = 334
5a = 524.9
a = 524.9/5
a = 104.98 (approx.)

Now we have found the values of a and d:
a ā‰ˆ 105 and d ā‰ˆ -19.09

Step 3: Finding the sum of the first twenty terms (S_20)
Using the formula for the sum of the first n terms, we can now find S_20:

S_20 = (20/2)(2a + (20-1)d)
S_20 = 10(2a + 19d)
Substituting the values of a and d:
S_20 = 10(2(105) + 19(-19.09))
S_20 = 10(210 - 363.71)
S_20 = 10(-153.71)
S_20 = -1537.1

Therefore, the sum of the first twenty terms of the AP is -1537.1.