Question

The sum of the 2nd term and the 7th term of an ap is 30. If its 15th term is 1 less than twice of its 8th term, find the ap

Answer 1

this is the answer to your questions

Answer 2

AP is 1, 5, 9, 13 if $\bold{15^{th}}$ term is 1 less than twice of its $\bold{8^{th}}$ term.

Solution:

Addition of $2^{nd}$ and $7^{th}$ term of AP = 30

(A+ D) + (A + 6D) = 30

Sum = 2A + 7D = 30.

Now as the question says the subtraction of 15th term and 2(8th term) = 1, that will be  

(A + 14D) – 2(A + 7D) = 1

(A + 14D) – 2(A + 7D) = 1

A – 2A = 1

Value of A = 1

First term = 1,  

Difference can be found by substituting value of A in (A+ D) + (A + 6D) = 30 ) we get,

(1+ D) + (1 + 6D) = 30

2 + 7D = 30

D = 4.  

With both the values of A and D, the Arithmetic Progression as 1, 5, 9, 13…..