# <span lang="EN-US

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 = 18th term, t19 = 19th term,

t20 = 20th term

We use nth term of an A.P. formula

tn = a + (n – 1)d

where n = no. of terms

a = first term

d = common difference

tn = nth 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, tn + 1 = tn + 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

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