an ap consists of 21 terms.the sum of 3terms in the middle is 129 and of the last three is 237 find the AP.
21terms/3terms=7groups of 3terms
237(last group)-129 (middle group)=36
237(group7)-36=201(group6)
group6-36
group5-36
129(group4)-36=95(group3)
group3-36
group2-36
add up all group terms(237+201+165...)
Hope that's what your looking for if not sorry, but that should be right
To find the arithmetic progression (AP), we need to determine the common difference (d) and the first term (a₁).
Let's start by finding the middle term of the AP. Since we have 21 terms, the middle term will be the 11th term.
Using the formula for the nth term of an AP:
Tn = a₁ + (n - 1) * d
Substituting n = 11, we have:
T11 = a₁ + (11 - 1) * d
T11 = a₁ + 10d
Now, let's find the sum of the three terms in the middle, which is 129. Since the middle term is the 11th term, the terms before it will be the 9th and 10th terms.
Sum of the three middle terms:
129 = T9 + T10 + T11
129 = a₁ + 8d + a₁ + 9d + a₁ + 10d
Simplifying the equation:
3a₁ + 27d = 129 ---(1)
Similarly, the sum of the last three terms is 237. Since the last term is the 21st term, the terms before it will be the 19th and 20th terms.
Sum of the last three terms:
237 = T19 + T20 + T21
237 = a₁ + 18d + a₁ + 19d + a₁ + 20d
Simplifying the equation:
3a₁ + 57d = 237 ---(2)
Now we have two equations:
(1) 3a₁ + 27d = 129
(2) 3a₁ + 57d = 237
To find the common difference (d) and the first term (a₁), let's subtract equation (1) from equation (2):
(2) - (1) gives:
30d = 108
Solving for d:
d = 108 / 30
d = 3.6
Substituting the value of d into equation (1) to find a₁:
3a₁ + 27(3.6) = 129
3a₁ + 97.2 = 129
3a₁ = 129 - 97.2
3a₁ = 31.8
a₁ = 31.8 / 3
a₁ = 10.6
Therefore, the first term (a₁) is 10.6, and the common difference (d) is 3.6.
The arithmetic progression (AP) is:
10.6, 14.2, 17.8, 21.4, 25, 28.6, 32.2, 35.8, 39.4, 43, 46.6, 50.2, 53.8, 57.4, 61, 64.6, 68.2, 71.8, 75.4, 79, 82.6