# 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 = 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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