Active forum topics
- Problems in progression
- Ceva’s Theorem Is More Than a Formula for Concurrency
- The Chain Rule Explained: Don't Just Memorize, Visualize It
- The Intuition Behind Integration by Parts (Proof & Example)
- Statics
- Calculus
- Hydraulics: Rotating Vessel
- Inverse Trigo
- Application of Differential Equation: Newton's Law of Cooling
- General Solution of $y' = x \, \ln x$
New forum topics
- Ceva’s Theorem Is More Than a Formula for Concurrency
- The Chain Rule Explained: Don't Just Memorize, Visualize It
- The Intuition Behind Integration by Parts (Proof & Example)
- Statics
- Calculus
- Hydraulics: Rotating Vessel
- Inverse Trigo
- Problems in progression
- General Solution of $y' = x \, \ln x$
- engineering economics: construct the cash flow diagram
Recent comments
- z3 days 19 hours ago
- Force P only impends motion…3 days 20 hours ago
- Wow! :>3 weeks ago
- In general, the centroid of …3 weeks 4 days ago
- isn't the centroid of the…3 weeks 4 days ago
- I get it now, for long I was…1 month 1 week ago
- Why is BD Tension?
is it not…1 month 1 week ago - Bakit po nagmultiply ng 3/4…3 months ago
- Determine the least depth…1 year 1 month ago
- Solve mo ang h manually…3 months ago


No it is not.
No it is not.
Arranged the numbers in
Arrange the numbers in increasing value: 1/6, 1/3, 1, 2. For a sequence to be arithmetic, the difference of two consecutive terms (any term - preceding term) must be equal. Let us check:
(a) 1/3 - 1/6 = 1/6
(b) 1 - 1/3 = 2/3
(c) 2 - 1 = 1
Since the difference of two adjacent terms are not equal, the given terms do not form into arithmetic sequence.
Arithmetic sequence must be
Arithmetic sequence must be in increasing and deceasing value.
No. The sequence must be 1/3,
No. The sequence must be 1/3, 1/6, 1, 2.
a, a+d, a+2d, a+3d. Where a = 1/3, and d = 1/3.