An AP consists of 37 terms. The sum of the three middle most terms is 225 and the sum of the last three terms is 429. Find the AP.

Given that, an AP has 37 terms.

⇒ n = 37 [∵ there are 37 terms ⇒ the last term is 37^th term]

We know middle term is given by,

Middle term of series = (n + 1)/2

= (37 + 1)/2

= 38/2

= 19^th term

If the mid-most term is at 19^th position, the three middle most terms will be at position 18^th, 19^th and 20^th.

Using relation, a_n = a + (n – 1)d …(i)

Where a = first term of the series,

n = position of the term

& d = common difference of the series

The term at 18^th position = a₁₈ = a + (18 – 1)d [Using equation (i), where n = 18]

⇒ a₁₈ = a + 17d …(ii)

The term at 19^th position = a₁₉ = a + (19 – 1)d [Using equation (i), where n = 19]

⇒ a₁₉ = a + 18d …(iii)

The term at 20^th position = a₂₀ = a + (20 – 1)d [Using equation (i), where n = 20]

⇒ a₂₀ = a + 19d …(iv)

According to the question, the sum of these three middle most terms = 225

⇒ (a + 17d) + (a + 18d) + (a + 19d) = 225 [Using equations (ii), (iii) & (iv)]

⇒ 3a + 54d = 225 …(v)

Also, if the last term is at 37^th position, the last three terms will be at 35^th, 36^th and 37^th position.

Similarly,

The term at 35^th position = a₃₇ = a + (35 – 1)d

⇒ a₃₅ = a + 34d …(vi)

The term at 36^th position = a₃₆ = a + (36 – 1)d

⇒ a₃₆ = a + 35d …(vii)

The term at 37^th position = a₃₇ = a + (37 – 1)d

⇒ a₃₇ = a + 36d …(viii)

According to the question, the sum of these last three terms = 429

⇒ (a + 34d) + (a + 35d) + (a + 36d) = 429

⇒ 3a + 105d = 429 …(ix)

Subtracting equation (v) from equation (ix), we get

(3a + 105d) – (3a + 54d) = 429 – 225

⇒ 105d – 54d = 204

⇒ 51d = 204

⇒ d = 204/51

⇒ d = 4

Putting d = 4 in equation (v), we get

3a + 54(4) = 225

⇒ 3a + 216 = 225

⇒ 3a = 225 – 216

⇒ 3a = 9

⇒ a = 3

The first term, a = 3

Common difference, d = 4

Last term, a₃₇ = a + 36d = 3 + 36(4) = 3 + 144 = 147

So, AP = 3, (3 + 4 =)7, (7 + 4 =)11, …, 147

Hence, AP = 3, 7, 11, … , 147