Question

an ap has 37 terms . the sum of the three middle most terms is225 and the sum of the last three terms is 429 .Find the AP

Answer 1

Let the first term and the common difference of the A.P are a and d respectively.

Since the A.P contains 37 terms.

So, the middle most term is (37+1)/2 th term = 19th term.

Thus, three middle most terms of this A.P.are 18th, 19th and 20th terms.

Given a 18 + a 19 + a 20 = 225
⇒ (a + 17d) + (a + 18d) + (a + 19d) = 225
⇒ 3(a + 18d) = 225
⇒ a + 18d = 75
⇒ a = 75 – 18d … (1)
According to given information

35 + a 36 + a 37 = 429
⇒ (a + 34d) + (a + 35d) + (a + 36d) = 429
⇒ 3(a + 35d) = 429
⇒ (75 – 18d) + 35d = 143
⇒ 17d = 143 – 75 = 68
⇒ d = 4
Substituting the value of d in equation (1), it is obtained
a = 75 – 18 × 4 = 3
Thus, the A.P. is 3, 7, 11, 15 …

Hope! This will help you.Cheers!

Answer 2

aₓ = a₁+(x-1)d
n=37
middle terms are a₁₈,a₁₉,a₂₀
a₁₈ + a₂₀ = 2a₁₉
a₁₈ + a₁₉ + a₂₀ = 225
a₁₉ = a₁ + 18d = 225/3 = 75
a₁ + 18d = 75..........................................................1
last terms are a₃₅,a₃₆,a₃₇
a₁ + 34d + a₁ + 35d +a₁ + 36d = 3a₁+105d=429
a₁ + 35d =143..........................................................2
17d = 68
∴d = 4  amd a₁ = 3

AP = 3, 7, 11,.......147