In an AP 4th and 7th terms are 17 and 23 respectively find 'd' and 'a

4th term of an AP = 17

7th term = 23

value of 'd' and 'a'

We know that,

nth term of an AP – a(n) = a + (n - 1)d

Hence,

a + (4 - 1)d = 17

→ a + 3d = 17 -- equation (1)

Similarly,

a + (7 - 1)d = 23

→ a + 6d = 23 -- equation (2)

Subtract equation (1) from (2).

→ a + 6d - (a + 3d) = 23 - 17

→ a + 6d - a - 3d = 6

→ 3d = 6

→ d = 6/3

→ d = 2

Substitute the value of d in equation (1).

→ a + 3(2) = 17

→ a + 6 = 17

→ a = 17 - 6

→ a = 11

Therefore,

I hope it will help you.

Regards.

Subtract equation (1) from (2).

→ a + 6d - (a + 3d) = 23 - 17

→ a + 6d - a - 3d = 6

→ 3d = 6

→ d = 6/3

→ d = 2

Substitute the value of d in equation (1).

→ a + 3(2) = 17

→ a + 6 = 17

→ a = 17 - 6

→ a = 11

Step-by-step explanation:

answer: standard form is ax^2 + bx + c = 0

step-by-step explanation:

standard form is ax^2 + bx + c = 0

(2x + 1)(x - 2) = 0

distribute to get into standard form

2x^2 - 4x + x - 2 = 0

2x^2 - 3x - 2 = 0

a = 2, b= -3, c = -2

a is your answer