Question

If an ap of 50 terms the sum of first terms is 210 and the sum of his last and 15 term is 2565 find the ip

Answer 1

Answer:the required AP is 3, 7, 11,....

Step-by-step explanation:

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,....

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Answer 2

$\underline{ \underline{\bf{Answer}}} : - \\ \implies \: 3, \: 7 \:, 11 \: ,15, \: ..........,199 \\ \\ \underline{\underline{ \bf{Step - by - step \: explanation \: }}} : -$

According to the question:-

$\bf{sum \: of \: first \: 10 \: terms \:( s_{10}) = 210} \\ 210 = \frac{10}{2} \bigg (2a + (101)d \bigg) \: \\ \\ 2a + 9d = 42 \: .........(1)\\ \\ \bf{sum \: of \: last \: 15 \: terms \: ( s_{15})= 2565} \\ \\ s_{50} -s_{35} = 2565 \\ \\ 2565 = \frac{50}{2} \bigg(2a + (50 - 1)d \bigg) - \frac{35}{2} \bigg(2a + (35 - 1)d \bigg) \\ \\ 2565 = 25(2a + 49d) - 35(a + 17d) \\ \\ 2565 = 50a + 1225d - 35a - 595d \\ \\ after \: solving \: this \: \\ \\ a + 42d = 171 \: ...........(2) \\ \\ from \: eq(1) \: and \: (2) \\ \\eq (1) \times 42 - \: eq (2) \times 9 \\ \\ we \: get \: \\ \\ a = 3 \: d = 4$

Hence required AP is →

3,7,11,15,....,199