Question

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

Answer 1

Since, total number of terms (n) = 37  ...[odd]

∴ Middle term = `((37 + 1)/2)^("th")` term = 19^th term

So, the three middle most terms = 18^th, 19^th and 20^th,

By given condition,

Sum of the three middle most terms = 225

a₁₈ + a₁₉ + a₂₀ = 225

⇒ (a + 17d) + (a + 18d) + (a + 19d) = 225   ...[∵ a_n = a + (n – 1)d]

⇒ 3a + 54d = 225

⇒ a + 18d = 75  ...(i)

And sum of the last three terms = 429

⇒ a₃₅ + a₃₆ + a₃₇ = 429

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

⇒ 3a + 105d = 429

⇒ a + 35d = 143   ...(ii)

On subtracting equation (i) from equation (ii), we get

17d = 68

⇒ d = 4

From equation (i),

a + 18(4) = 75

⇒ a = 75 – 72

⇒ a = 3

∴ Required AP is a, a + d, a + 2d, a + 3d,...

i.e., 3, 3 + 4, 3 + 2(4), 3 + 3(4),...

i.e., 3, 7, 3 + 8, 3 + 12,...

i.e., 3, 7, 11, 15,...