The second term of an arithmetic progression I four times the fits term, and the first term is ten.

Find the common difference and hence find the sum of the first 12 terms.

The sum of the 3rd and 7th term of an AP is 38 and the 9th term is 37.find the AP

To find the common difference of an arithmetic progression (AP), we can use the formula:

nth term = first term + (n - 1) * common difference

We are given that the second term is four times the fifth term, so we can write:

2nd term = 4 * 5th term

Plugging in the values, we get:

10 + (2 - 1) * common difference = 4 * (10 + (5 - 1) * common difference)

10 + common difference = 4 * (10 + 4 * common difference)

10 + common difference = 4 * 10 + 16 * common difference

10 - 40 = 16 * common difference - common difference

-30 = 15 * common difference

common difference = -30 / 15

common difference = -2

So, the common difference of the arithmetic progression is -2.

To find the sum of the first 12 terms of the arithmetic progression, we can use the formula:

Sum = (n / 2) * (2 * first term + (n - 1) * common difference)

Plugging in the values, we get:

Sum = (12 / 2) * (2 * 10 + (12 - 1) * -2)

Sum = 6 * (20 + 11 * -2)

Sum = 6 * (20 - 22)

Sum = 6 * -2

Sum = -12

Therefore, the sum of the first 12 terms of the arithmetic progression is -12.

To find the common difference, let's use the given information: the second term of the arithmetic progression is four times the fifth term, and the first term is ten.

Let's assume that the common difference is denoted by 'd'.

Given:
Second term = 4 times the fifth term
First term = 10

The formula for the n-th term of an arithmetic progression is:
an = a1 + (n - 1)d,

where 'an' is the n-th term, 'a1' is the first term, 'n' is the position of the term in the series, and 'd' is the common difference.

Using the given information, we can form the equations:

a2 = 4a5, and a1 = 10.

Substituting these values into the formula:

a2 = a1 + (2 - 1)d,
4a5 = a1 + (5 - 1)d.

Plugging in the given values:

10 + d = 4(10 + 4d),
40d = 9d.

Now we can solve this equation for 'd':

10 + d = 40 + 16d,
30d = -30,
d = -1.

The common difference is -1.

To find the sum of the first 12 terms, we can use the formula for the sum of an arithmetic series:

Sn = (n/2)(2a1 + (n - 1)d).

Plugging in the known values:

S12 = (12/2)(2(10) + (12 - 1)(-1)),
S12 = 6(20 + 11),
S12 = 6(31),
S12 = 186.

Therefore, the sum of the first 12 terms of the arithmetic progression is 186.

a+d = 4a

a = 10

10+d = 40
d = 30

S12 = 12/2 (2*10 + 11*30)