let the first 4 terms of the AP be
a-d, a , a+d, and a+2d
new sequence is
a-d+1, a+2, a+d + 7, and a+2d+20
so (a+2)/(a-d+1) = (a+d+7)/(a+2)
(a+2)^2 = (a-d+1)(a+d+7)
a^2+4a+4 = a^2+ad+7a-ad-d^2-7d+7a+d+7
-4a = -d^2 - 6d + 3
4a = d^2 + 6d - 3 **
(a+d+7)/(a+2) = (a+2d+20)/(a+d+7)
(a+d+7)^2 = (a+2)(a+2d+20)
a^2 + d^2 + 49 + 2ad +14a + 14d
= a^2 + 2ad + 22a + 4d + 40
8a = d^2 + 10d + 9 ***
double **, then subtract ***
0 = d^2 + 2d - 15 = 0
(d + 5)(d-3) = 0
d = -5, or d = 3
if d = 3 in **
4a = 3^2 + 18 - 3 = 24
a = 6
the original AP according to my definition , was 3, 6, 9, 12
check: if I add the numbers as stated, I get 4, 8, 16, and 32 , which is a GP
if d = -5, in **
4a = 25 -30 - 3 = -8
a = -2
the original AP is 3, -2, -7, -12
adding the numbers as stated will give me:
4, 0, 0, 8 , but we can't have a 0 in a GP
so we have to go with my first part of the solutions, which yielded the AP
3, 6, 9, 12
Making the first term 3, and the common difference as 3
if 1, 2, 7 and 20, respectively, are added to the first terms of an arithmetic progression, a geometric progression of four terms is obtained. find the first term and common difference of the arithmetic progression
the answers are both 3 .. but i don't know the solution, please help me, thank you
1 answer
1 answer