Firstly an AP, some minus becomes GP!
4 positive integers form an arithmetic progression.
If we subtract 2 , 6 , 7 and 2 , respectively, from the 4 numbers, it forms a geometric progression.
What is the sum of these 4 numbers?
The answer is 62.
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2 solutions
Also, since the AP consists of positive integers, a and r are both integers. Thus from (1) we directly have a = 3 and ( r − 1 ) 2 = 1 ⇒ r = 2
did the same way!
Well , I did it the long way . let us say the AP is a , a + d , a + 2 d , a + 3 d . Now a − 2 , a + d − 6 , a + 2 d − 7 , a + 3 d − 2 are in GP ,That is ( a + d − 6 ) 2 = ( a + 2 d − 7 ) ( a − 2 ) a n d ( a + 2 d − 7 ) 2 = ( a + 3 d − 2 ) ( a + d − 6 ) . On solving we get a = 5 , d = 1 o r a = 5 , d = 7 but d = 1 makes second term of the GP zero which is not possible . Hence the no. are 5 , 1 2 , 1 9 , 2 6
Did the same! What's the short way?
I did it backwards...
Let the GP is a , a r , a r 2 , a r 3 .
⇒ a + 2 , a r + 6 , a r 2 + 7 , a r 3 + 2 form an AP.
Thus, ( a r + 6 ) − ( a + 2 ) = ( a r 2 + 7 ) − ( a r + 6 )
⇒ a ( r − 1 ) 2 = 3 . . . ( 1 )
Also, ( a r 2 + 7 ) − ( a r + 6 ) = ( a r 3 + 2 ) − ( a r 2 + 7 )
⇒ a r ( r − 1 ) 2 = 6 . . . ( 2 )
Dividing (2) by (1), we get
r = 2 ⇒ a = 3
Thus, the AP is 5, 12, 19, 26.
And answer = 5 + 1 2 + 1 9 + 2 6 = 6 2