Question

please solve it fastly it is very urgent ​

Answer 1

Answer:

3 7 11 15 19 23 27 31 35 39 .......

Answer 2

Answer:

AP = 3, 7, 11, 15, 19, 23, 27,......

Step-by-step explanation:

Let the 1st term be a and their common difference be d.

Now, we are given

S(10) = 210

and Sum of last 15 terms = 2565

Now, to find the sum of n terms of an AP we use the general formula,

Sn = (n/2)[2a + (n - 1)d]

Thus,

here n = 10

S(10) = (10/2)[2a + (10 - 1)d]

210 = 5 × (2a + 9d)

2a + 9d = 210/5

2a + 9d = 42 ----- 1

Now,

To get the sum of the last 15 terms,

We must subtract sum of 50 terms with sum of 35 terms, because 50 - 15 = 35,

Thus,

S(50) - S(35) = 2565

Now,

S(50) = (50/2)[2a + (50 - 1)d]

S(50) = 25(2a + 49d)

S(50) = 50a + 1225d

Also,

S(35) = (35/2)[2a + (35 - 1)d]

S(35) = (35/2)(2a + 34d)

Taking 2 as common factor,

S(35) = (35/2) × 2(a + 17d)

S(35) = 35 × (a + 17d)

S(35) = 35a + 595d

Thus,

S(50) - S(35) = 2565

(50a + 1225d) - (35a + 595d) = 2565

Opening the brackets,

50a + 1225d - 35a - 595d = 2565

15a + 630d = 2565

Taking 15 as common factor,

15(a + 42d) = 2565

a + 42d = 2565/15

a + 42d = 171

Multiplying the whole equation by 2,

2a + 84d = 342 ----- 2

Subtracting eq.1 in eq.2 we get,

(2a + 84d) - (2a + 9d) = 342 - 42

Opening the brackets,

2a + 84d - 2a - 9d = 300

75d = 300

d = 300/75

d = 4

Putting d = 4 in eq.1 we get,

2a + 9(4) = 42

2a + 36 = 42

2a = 42 - 36

2a = 6

a = 6/2

a = 3

Hence,

1st term is 3 and their common difference is 4

Thus,

AP = 3, 7, 11, 15, 19, 23, 27,......

Hope it helped and believing you understood it........All the best