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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.

Print or Save**Sol**. 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

\(({{37+1}\over2})\)^{th} term = 19^{th} term

Thus three middle most term of this

A.P are 18^{th}, 19^{th} and 20^{th} terms.

t_{18} + t_{19} + t_{20} = 225

a + 17d + a + 18d + a + 19d = 225

[ tn = a + (n –1)d]

3a + 54d = 225

(Dividing both sides by 3 throughout)

According to condition (II);

a + 18d = 75 ...(I)

Now, t_{35} + t_{36} + t_{37} = 429

a + 34d + a + 35d + a + 36d = 429

[tn = a + (n –1)d]

3a + 105d = 429

(Dividing both sides by 3 throughtout)

a + 35d = 143 ...(II)

Subtracting (I) from (II)

a + 35d = 143

– a + 18d = 75

(–) (–) (–)

17d = 68

d = 4

Substituting d = 4, in equation (I)

a + 18 × 4 = 75

a + 72 = 75

a = 75 – 72

a = 3

∴ t_{1} = a = 3

t_{2} = t1 + d = 3 + 4 = 7

t_{3} = t2 + d = 7 + 4 = 11

**∴ The Required AP is 3, 7, 11, ...**