Question

All terms of arithmetic sequence is natural numbers its common difference is 3 find its sum of reciprocal of two consecutive terms of arithmetic sequence is 11 ÷28 find the terms​

Answer 1

Let the first term of the A.P. be a

Then the A.P. can be written as

a, a + 3, a + 6, a + 9, ...

Given that, the sum of reciprocal of two consecutive terms is 11/28.

• Then, 1/a + 1/(a + 3) = 11/28

• or, (a + 3 + a)/{a (a + 3)} = 11/28

• or, 28 (2a + 3) = 11a (a + 3)

• or, 56a + 84 = 11a² + 33a

• or, 11a² - 23a - 84 = 0

• or, 11a² - 44a + 21a - 84 = 0

• or, 11a (a - 4) + 21 (a - 4) = 0

• or, (a - 4) (11a + 21) = 0

• This gives a = 4, - 21/11

Since a cannot be negative or fraction, the value of a is 4.

Therefore the required A.P. is

4, 7, 10, 13, 16, 19, ...