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In an AP of 50 terms, the sum of 10 terms is 210 and the sum of last 15 terms is 2565 . Find the AP

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Answered by Anonymous
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\mathcal{\huge{\underline{\underline{\red{Question ?}}}}}

In an AP of 50 terms, the sum of 10 terms is 210 and the sum of last 15 terms is 2565 . Find the AP.

\mathcal{\huge{\underline{\underline{\green{Answer:-}}}}}

✒ The AP is 3, 7, 11, 15, 19, 23, 27, 31, 35, 39 ........

\mathcal{\huge{\underline{\underline{\blue{Solution:-}}}}}

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

-      -         -

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        75d = 300

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 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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