Question

Is an ap of 50 terms the sum of the first 10 term is 210 and the sum of first last 15 terms is 2 and 2565 find the ap?

Answer 1

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Let the first term be (a) and (d) be the common difference .
=) Sum of the first n terms is given by
Sn = n/2 {2a + (n - 1)d}
On Putting n = 10 =)
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) we get :-
 
 =)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 .

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