the 1st term of arithmetic series is 2 .the 1st ,3rd and 11th terms are the first the terms of a geometric series. find the 6 terms of the arithemetic series

To find the 6 terms of the arithmetic series, we need to determine the common difference (d) first.

We know that the 1st term (a1) is 2. Let's assume the common difference is d.

Thus, the 2nd term (a2) would be:
a2 = a1 + d = 2 + d

The 3rd term (a3) would be:
a3 = a1 + 2d = 2 + 2d

And the 11th term (a11) would be:
a11 = a1 + 10d = 2 + 10d

We are also given that the 1st, 3rd, and 11th terms of the arithmetic series form the first terms of a geometric series. This means that the ratio between consecutive terms in the geometric series is constant. Let's assume this ratio is r.

Thus, the 2nd term of the geometric series would be:
a2 = a1 * r = 2 * r

The 3rd term of the geometric series would be:
a3 = a2 * r = 2 * r * r = 2r^2

And the 11th term of the geometric series would be:
a11 = a1 * r^10 = 2 * r^10

Since we know the 1st, 3rd, and 11th terms of the geometric series, we can create two equations:

1st Arithmetic term = 1st Geometric term
2 + d = 2r

3rd Arithmetic term = 3rd Geometric term
2 + 2d = 2r^2

From the first equation, we can express d in terms of r:
d = 2r - 2

Substituting this value of d into the second equation, we get:
2 + 2(2r - 2) = 2r^2
2 + 4r - 4 = 2r^2
4r - 2 = 2r^2
2r^2 - 4r + 2 = 0

Now we can solve this quadratic equation for r using the quadratic formula:
r = (-b ± √(b^2 - 4ac)) / (2a)

Plugging in the values for a, b, and c:
r = (4 ± √((-4)^2 - 4(2)(2))) / (2(2))
r = (4 ± √(16 - 16)) / 4
r = (4 ± √0) / 4

Since the discriminant (√(b^2 - 4ac)) is 0, there is only one real solution for r.
r = 4 / 4
r = 1

Now that we have determined the common ratio (r = 1), we can find the value of d using the previous equation:
d = 2r - 2
d = 2(1) - 2
d = 0

Since d = 0, the arithmetic series is actually a constant series with all terms equal to 2. Therefore, the 6 terms of the arithmetic series are:
2, 2, 2, 2, 2, 2

To find the 6th term of the arithmetic series, we need to determine the common difference of the series.

Let's start by finding the common difference using the 1st, 3rd, and 11th terms of the arithmetic series.

The 1st term of the arithmetic series is given as 2.

The 3rd term, denoted as a3, can be calculated using the formula: a3 = a1 + (3 - 1)d, where a1 is the 1st term and d is the common difference. Substituting the values, we have a3 = 2 + 2d.

Similarly, the 11th term, denoted as a11, can be calculated using the formula: a11 = a1 + (11 - 1)d, which becomes a11 = 2 + 10d.

Now let's examine the geometric series formed by the 1st, 3rd, and 11th terms.

In a geometric series, each term is obtained by multiplying the previous term by a common ratio.

So, the common ratio, denoted as r, can be calculated by dividing the 3rd term (a3) by the 1st term (a1) of the geometric series: r = a3 / a1 = (2 + 2d) / 2 = 1 + d.

Similarly, the 11th term to the 3rd term ratio, denoted as r', can be calculated as: r' = a11 / a3 = (2 + 10d) / (2 + 2d) = 1 + 4d.

Now we equate both ratios to find the value of d:

1 + d = 1 + 4d

By solving this equation, we find that d = 1/3.

Now that we have the common difference, we can find the 6 terms of the arithmetic series by using the formula:

an = a1 + (n - 1)d, where an is the nth term.

Substituting the values, we get:

a1 = 2 (given)
d = 1/3 (calculated)

a2 = a1 + (2 - 1)d = 2 + (1/3) = 7/3
a3 = a1 + (3 - 1)d = 2 + (1/3)(2) = 8/3
a4 = a1 + (4 - 1)d = 2 + (1/3)(3) = 11/3
a5 = a1 + (5 - 1)d = 2 + (1/3)(4) = 14/3
a6 = a1 + (6 - 1)d = 2 + (1/3)(5) = 17/3

Therefore, the 6 terms of the arithmetic series, starting from the 1st term, are:
2, 7/3 , 8/3 , 11/3 , 14/3 , 17/3

if the GP is a,ar,...

2 = a
2+2d = ar
2+10d = ar^2
a=2,d=3,r=4

AP: 2,5,8,11,14,17,20,23,26,29,32,...
GP: 2,8,32,...