Question

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

Answer 1

Addition of and 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…..