the sum of the 4th and 6th terms of an A.P is 42. the sum of the 3rd and 9th terms of the progression is 52. find the first term, the common difference and the sum of the first ten terms of the progression.

what don't you get?

T4 = a+3d
T6 = a+5d
a+3d + a+5d = 42
2a + 8d = 42
a+4d = 21

T3 = a+2d
T9=a+8d
a+2d + a+8d = 52
2a + 10d = 52
a+5d = 26

a+4d = 21
a+5d = 26
d = 5
so a=1

sum of T1-T10 = 10/2(1+46) = 235

AP = 1 6 11 16 21 26 31 36 41 46 . . .

Answer

I need freaking help right now and this is not helping me!!!!!

it is not helping matters

To solve this problem, we'll use the formula for the nth term of an arithmetic progression (A.P) and the formula for the sum of the first n terms of an A.P.

Let's start by finding the common difference (d) of the A.P:

Given that the sum of the 4th and 6th terms is 42, we can write this as:

a4 + a6 = 42

Using the formula for the nth term of an A.P, we know that:

a4 = a1 + 3d (since a4 is the first term plus 3d)
a6 = a1 + 5d (since a6 is the first term plus 5d)

Substituting these expressions into the equation, we have:

(a1 + 3d) + (a1 + 5d) = 42
2a1 + 8d = 42 -------- (Equation 1)

Next, let's find the common difference (d) using the sum of the 3rd and 9th terms:

Given that the sum of the 3rd and 9th terms is 52, we can write this as:

a3 + a9 = 52

Using the same logic as before, we get:

a3 = a1 + 2d (since a3 is the first term plus 2d)
a9 = a1 + 8d (since a9 is the first term plus 8d)

Substituting these expressions into the equation, we have:

(a1 + 2d) + (a1 + 8d) = 52
2a1 + 10d = 52 --------- (Equation 2)

Now, we have a system of two equations with two variables (a1 and d):
Equation 1: 2a1 + 8d = 42
Equation 2: 2a1 + 10d = 52

We can solve this system of equations to find the values of a1 and d.

Subtracting Equation 1 from Equation 2, we get:

2a1 + 10d - (2a1 + 8d) = 52 - 42
2a1 + 10d - 2a1 - 8d = 10
2d = 10
d = 10/2
d = 5

Now that we have the value of d, let's substitute it back into Equation 1 to find a1:

2a1 + 8(5) = 42
2a1 + 40 = 42
2a1 = 42 - 40
2a1 = 2
a1 = 2/2
a1 = 1

Therefore, the first term (a1) of the A.P is 1, and the common difference (d) is 5.

To find the sum of the first ten terms of the progression, we can use the formula for the sum of the first n terms of an A.P:

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

Substituting the values we found:

S10 = 10/2 * (2(1) + (10 - 1)(5))
S10 = 5 * (2 + 9(5))
S10 = 5 * (2 + 45)
S10 = 5 * 47
S10 = 235

Therefore, the sum of the first ten terms of the A.P is 235.