Question

The first, second and fifth terms of an arithmetic progression are the first three terms of a geometric progression. The third term of the arithmetic progression is 5. Find the 2 possible values for the fourth term of the geometric progression.

Answer 1

ANSWER
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a1∗a13=a3∗a3a1∗a13=a3∗a3
and
a3+a7=40.a3+a7=40.



Therefore a3=a1+2d,a3=a1+2


, a13=a1+12da13=a1+12d, a4=a1+3da4=a1+3d, a7=a1+6da7=a1+6d,


we get
a1(a1+12d)=(a1+2d)2a1(a1+12d)=(a1+2d)2

and a1+3d+a1+6d=40.a1+3d+a1+6d



, 12a1d=4a1d+4d212a1d=4a1d+4d2 or 8a1d=4d28a1d=4d2 which leads to 2a1=d.2a1=d.



...... =
The other equation leads to 2a1+9d=402a1+9d=40, so 2a1+18a1=40,2a1+18a1=40, or

a1=2.a1=2.



Therefore the answer is 2

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