Sequence And Series Question 519
Three numbers are in A.P. such that their sum is 18 and sum of their squares is 158. The greatest number among them is
[UPSEAT 2004]
Options:
10
11
12
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
Let three number of A.P. $ a-d,,a,, $ $ a+d $ Sum = 18, and $ {{(a-d)}^{2}}+a^{2}+{{(a+d)}^{2}}=58 $ $ a-d+a+a+d=18 $ $ a=6 $ and $ {{(6-d)}^{2}}+36+{{(6+d)}^{2}}=158 $ = $ 36-2d^{2}+36+d^{2}=122 $ $ =2d^{2}+72=122 $ $ =2d^{2}=50 $ Þ $ d=5 $ . Hence Numbers are 1, 6, 11, i.e. maximum number is 11.
BETA
