The sum of the 5th term of the 7th term of an ap is 52 and the 10th term is 46 find the ap

The AP is, 1 , 6, 11, 16, 21, 26, 31,

Step-by-step explanation:

From the question;

We are required to find the AP;

nth term = a + (n-1) d , where a is the first term and d is the common difference

5th term = a + 4d

7th term = a + 6d

5th term + 7th term = 2a + 10d = 52  eqn 1

10th term = a + 9d

2a + 10d = 52

a + 9d = 46

Multiplying the second equation by 2, we get;

2a + 10d = 52

2a + 18d = 92

Eliminating a by subtracting the two equations; we get;

-8d = -40

   d = 5

Solving for a

a + 9d = 46

   a = 46 - 9d

      = 46 - 9(5)

      = 46 - 45

   a  = 1

Thus, the first term, a= 1 and the common difference, d= 5

Therefore;

First term = a = 1

Second term = a + d = 1 + 5 = 6

Third term = a + 2d = 1 + 2(5) = 11

Fourth term = a + 3d = 1 + 3(5) =16

Fifth term = a + 4d = 1 + 4(5) = 21

Sixth term = 1 + 5d = 1 + 5(5) = 26

Seventh term = 1 + 6d = 1 + 6(5) = 31

Thus,

The AP is, 1 , 6, 11, 16, 21, 26, 31,

d) the triangle is a right triangle 60 squared +144 squared= 156 squared

step-by-step explanation:

156 squared is the hypothenuse

so 156= c