Sequence And Series Question 212
Question: Four numbers are in arithmetic progression. The sum of first and last term is 8 and the product of both middle terms is 15. The least number of the series is
[MP PET 2001]
Options:
A) 4
B) 3
C) 2
D) 1
Show Answer
Answer:
Correct Answer: D
Solution:
Let $ A_1,A_2,A_3 $ and $ A_4 $ are four numbers in A.P. $ A_1+A_4=8 $ ?..(i) and $ A_2.,A_3=15 $ ?..(ii) The sum of terms equidistant from the beginning and end is constant and is equal to sum of first and last terms. Hence, $ A_2+A_3=A_1+A_4=8 $ ?..(iii) From (ii) and (iii), $ A_2+\frac{15}{A_2}=8 $
Þ $ A_2^{2}-8A_2+15=0 $ $ A_2=3or5 $ and $ A_3=5,or3 $ . As we know, $ A_2=\frac{A_1+A_3}{2} $
Þ $ A_1=2A_2-A_3 $
Þ $ A_1=2\times 3-5=1 $ and $ A_4=8-A_1=7 $ Hence the series is, 1, 3, 5, 7. So that least number of series is 1.
BETA
