There are 37 terms in an A.P., the sum of three terms placed exactly at the middle is 225 and the sum of last three terms is 429. Write the A.P.

Let first term = a

Common difference = d

Since, A.P. consist of 37 terms, therefor the middle most term is

Thus, three middle most term are t₁₈ = 18^th term, t₁₉ = 19^th term,

t₂₀ = 20^th term

We use n^th term of an A.P. formula

t_n = a + (n – 1)d

where n = no. of terms

a = first term

d = common difference

t_n = n^th terms

Thus, on substituting all values we get,

Given, t₁₈ + t₁₉ + t₂₀ = 225

⇒ [a + (18 – 1)d] + [a + (19 – 1)d] + [a + (20 – 1)d] = 225

⇒ [a + 17d] + [a + 18d] + [a + 19d] = 225

⇒ 3a + 54d = 225

Dividing by 3

⇒ a + 18d = 75 …….(1)

Given, sum of last three term is 429

⇒ t₃₅ + t₃₆ + t₃₇ = 429

⇒ [a + (35 – 1)d] + [a + (36 – 1)d] + [a + (37 – 1)d] = 429

⇒ [a + 34d] + [a + 35d] + [a + 36d] = 429

⇒ 3a + 105d = 429

Dividing by 3

⇒ a + 35d = 143 …….(2)

Subtracting eq. (1) from eq. (2) we get,

⇒ [a + 35d] – [a + 18d] = 143 – 75

⇒ 17d = 68

Substituting value of ‘d’ in eq. (1) we get,

⇒ a + 18 × 4 = 75

⇒ a + 72 = 75

⇒ a = 75 – 72 = 3

⇒ a = t₁ = 3

We know that, t_{n + 1} = t_n + d

t₂ = t₁ + d = 3 + 4 = 7

t₃ = t₂ + d = 7 + 4 = 11

t₄ = t₃ + d = 11 + 4 = 15

t₃₇ = 3 + (37 – 1) × 4

t₃₇ = 3 + 36 × 4

t₃₇ = 3 + 144 = 147

Thus, the A.P. is 3, 7, 11, . . . . ., 147