What is the 260th number in the following sequence (1/1, 1/3, 2/3, 3/3, 1/5, 2/5, 3/5, 4/5, 5/5, 1/7, 2/7, …)? | Arithmetical And Geometrical Sequence | Arithmetic Sequence Is One in Which the Difference of Consecutive Terms is a Constant

 What is the 260th number in the following sequence (1/1, 1/3, 2/3, 3/3, 1/5, 2/5, 3/5, 4/5, 5/5, 1/7, 2/7, …)?

There are 2 types of sequence: i.e arithmetical and geometrical sequence.

    An arithmetic sequence is one in which the difference of consecutive terms is a constant. We often symbolize this constant difference by d, and therefore to determine the general (nth) term of an arithmetic sequence simply use this formula.

an=a1+(n-1)d , where,

a1 is the first term

n is the term, and

d is the common difference.

while;

A geometric sequence is one in which the ratio of consecutive terms is a constant. We often symbolize this constant ratio by r, and therefore to determine the general (nth) term of a geometric sequence just use the following formula

an=a1r^(n-1) where,

a1 is the first term

r is the common ratio an

n is the term.

Now back to yo question I will choose any two pairs of consecutive terms and prove wether they have common d or r.

therefore, (5/5, 1/7) and (3/3, 1/5).

start by checking wether it has common difference d .

5/5–1/7= 6/7 and 3/3–1/5= 4/5 hence not arithematic.

lets prove the same terms if they have common ratio.

1/7÷5/5=1/7 and 1/5 ÷ 3/3=1/5 hence no gemeotric.

then your terms are consecutively invalid!