Question

Q18 In an AP of 50 terms the
sum of first 10 term is210
and sum of its last 15 terms is 2565 . Find the AP
​

Answer 1

Answer:

Let a be the first term and d be the common difference of the given AP.

Sum of the first n terms is given by

Sn = n/2 {2a + (n - 1)d}

Putting n = 10, we get

S₁₀ = 10/2 {2a + (10 - 1)d}

210 = 5 (2a + 9d)

2a + 9d = 210/5

2a + 9d = 42 ...............(1)

Sum of the last 15 terms is 2565

⇒ Sum of the first 50 terms - sum of the first 35 terms = 2565

S₅₀ - S₃₅ = 2565

⇒ 50/2 {2a + (50 - 1)d} - 35/2 {2a + (35 - 1)d} = 2565

25 (2a + 49d) - 35/2 (2a + 34d) = 2565

⇒ 5 (2a + 49d) - 7/2 (2a + 34d) = 513

⇒ 10a + 245d - 7a + 119d = 513

⇒ 3a + 126d = 513

⇒ a + 42d = 171 ........(2)

Multiply the equation (2) with 2, we get

2a + 84d = 342 .........(3)

Subtracting (1) from (3)

2a + 84d = 342

2a + 9d = 42

- - -

_______________

75d = 300

_______________

d= 4

Now, substituting the value of d in equation (1)

2a + 9d = 42

2a + 9*4 = 42

2a = 42 - 36

2a = 6

a = 3

So, the required AP is 3, 7, 11, 15, 19, 23, 27, 31, 35, 39 ........

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