Question

the four of an ap is equal to 3 times the first one and a 7 exceeds twice the third term by 1 find s15​

Answer 1

Answer:

S15 = 255

Step-by-step explanation:

first term of AP is = a

4th term of AP is = a + 3d

4th term is 3 times of first term

a + 3d = 3a

a - 3a + 3d = 0

-2a + 3d = 0 equ (1)

3rd term = a + 2d

7th term = a + 6d

a + 6d = 2(a + 2d) + 1

a + 6d = 2a + 4d + 1

a + 6d - 2a - 4d = 1

-a + 2d = 1 equ (2)

multiply equ (2) with 2

- 2a + 4d = 2 equ (3)

subtract equ (3) - (1)

we get d = 2

substitute d = 2 in equ (2)

- a + 2d = 1

- a + 2(2) = 1

- a + 4 = 1

- a = 1 - 4

- a = - 3

a = 3

S15 = n/2 (2a + (n-1)d)

$= \frac{15}{2} (2(3) + (15 - 1)2) \\ = \frac{15}{2} (6 + 28) \\ = \frac{15}{2} \times 34 \\ = 15 \times 17 \\ = 255$

sum of 15 terms is 255