Q) In an A.P. of 50 terms, the sum of first 10 terms is 210 and the sum of last 15 terms is 2565. Find the A.P.

Ans:

We can find the AP, if we find its first term and common difference.

Let’s consider the first terms of AP is “a” and the common difference is “d”.

Step 1: Since the sum of first n terms of an AP, S_n = [2 a + (n – 1) d]

∴ Sum of first 10 terms, S₁₀ = [2 a + (10 – 1) (d)] = 5 (2 a + 9 d)

Step 2: Since it is given that the sum of first 10 terms is 210

∴ 5 (2 a + 9 d) = 210

∴ 2 a + 9 d = 42 …… (i)

Step 3: Next, it is given that the sum of last 15 terms is 2565

Since, Sum of last 15 terms = Sum of first 50 terms – Sum of first 35 terms

∴ S₅₀ – S₃₅ = 2565

∴ [2 a + (50 – 1) (d)] – [2 a + (35 – 1) (d)] = 2565

∴ 25 (2 a + 49 d) –  (2 a + 34 d) = 2565

∴ 25 (2 a + 49 d) –  35 (a + 17 d) = 2565

∴ 5 (2 a + 49 d) –  7 (a + 17 d) = 513

∴ 10 a + 245 d – 7 a – 119 d = 513

∴ 3 a + 126 d = 513

∴ a + 42 d  = 171 ….. (ii)

Step 4: By multiplying equation (ii) by 2 and then subtracting equation (i) from it, we get:

2 (a + 42 d) – (2 a + 9 d) = 2 (171 – 42

∴ 2 a + 84 d – 2 a – 9 d = 342 – 42

∴ + 75 d – – 9 d = 300

∴ 75 d  = 300

∴ d = 

∴ d = 4

Step 5: By substituting value of d in equation (i), we get:

2 a + 9 d = 42

∴ 2 a + 9 (4) = 42

∴ 2 a + 36 = 42

∴ 2 a = 42 – 36 = 6

∴ a =

∴  a = 3

Step 6: Let’s calculate the value of last term of the AP:

Since the value of n^th term, T_n = a + (n – 1) d

∴ T₅₀ = 3 + ( 50 – 1) (4)

∴ T₅₀ = 3 + 49 (4) = 3 + 196

∴ T₅₀ = 199

Therefore, the AP is 3, 7, 11,15,19, ……..199.

[Note: Here, since it is given that the AP has 50 terms it means it is definite AP and hence, we can leave our answer open ended like 3, 7, 11, …… Since it has 50 terms, we need to mention the value of 50th term when we write the AP.]

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