Question

the sum of 2nd and 7th terms of am arithmetic progression is 30. if 8ts 15th term is 1 less than twice it's 8th term. find the arithmetic progression​

Answer 1

Answer:

Sum = 2A + 7D = 30.

(A+ D) + (A + 6D) = 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…..

also ,this may help

Answer 2

Let,first term of AP be a  and common difference be d

Then,

      2nd term of AP=a+d

       7th term of AP=a+6d

sum=a+d+a+6d

30=2a+7d

2a+7d=30 --------- (eq-1)

8th term of AP=a+7d

15th term of AP=a+14d

According to question:

                          => a+14d+1=2(a+7d)

                           =>a+14d+1=2a+14d

                            =>a+14d-2a-14d=-1

                                 => -a=-1

                                   a=1

Putting value of a in  (eq-1):

2*1+7d=30

2+7d=30

7d=30-2

7d=28

d=28/7

d=4

So,Arithemetic Progression is :

= a,a+d,a+2d,a+3d.........

= 1,5,9,13.............