Quiz
Arithmetic & Geometric
Progressions
Let the first
term of the arithmetic progression be a and common difference be d.
First term = a
Third term = a + (3 - 1)
d = a + 2d
Thirteenth term = a + (13 - 1) d = a + 12d
Since the first, third and thirteenth terms are in geometric progression,
(a + 2d)2 = a (a + 12d)
a2 + 4ad + 4d2 = a2 + 12ad
2ad - d2 =
0
.....(1)
Fourth term + Seventh term = 40
a + (4 - 1)d + a (7 - 1)d = 40
2a + 9d = 40
2a = 40 - 9d .....(2)
substitute (2) into (1):
d (40 - 9d) - d2 = 0
d (4 - d) = 0
d = 0 (rejected) or d = 4
a = 2
Let a be
the first term of arithmetic progression, d be common difference and r
be common ratio of the geometric progression.
First term of G. P. = 9th
term of A. P.
= a + 8d
Sum to infinity = 80
a + 8d = 80 - 8r .....(1)
(a + 12d)2 = (a + 8d) (a
+14d) geometric
mean
d (a + 16d) = 0
d = 0 (rejected) or a =
-16d .....(2)
Given 16th term of A. P. = 4th term of G. P.,
a + 15d = (a + 8d) r3
-16d + 15d = (-16d + 8d) r3 subst (2)
r3 = 1/8
r = 1/2
-8d = 80 -
8r substitute (2) into (1)
d = -(10 -
8(1/2)) substitute
r = 1/2
= -5
a = 80
Sum of 16 terms = 16/2 [2(80) + 15(-5)]
= 680
Let a be
the first term and r be common ratio of G. P.
Sum of first 4 terms,
Sum of first 2 terms,
since
r>0
Fourth term,
a = 6
Sum to infinity,
Sum to first n terms
Least value of n is 19.