the 4th term of an ap is equal to 3 times the first term and the 7th term exceeds twice the third term by 1 find the first term and the common difference

We're given

a₄ = 3a₁

a₇ = 2a₃ + 1

Since these numbers form an arithmetic progression, we have for some fixed constant d

a₂ = a₁ + d

a₃ = a₂ + d = a₁ + 2d

a₄ = a₃ + d = a₁ + 3d

and so on, up to

a₇ = a₁ + 6d

Rewrite the given equations in terms of a₁, then solve for a₁ and d :

a₁ + 3d = 3a₁

a₁ + 6d = 2 (a₁ + 2d) + 1

2a₁ - 3d = 0

a₁ - 2d = -1

(2a₁ - 3d) - 2 (a₁ - 2d) = 0 - 2 (-1)

2a₁ - 3d - 2a₁ + 4d = 0 + 2

d = 2

a₁ + 3d = 3a₁

3 (2) = 2a₁

a₁ = 3

1/3 + 1/3 ≠ 2/6 because 1/3 alone = 2/6  

step by step explanation:

1/3 + 1/3 = 2/3       2/3=4/6

answer 1 : option "c"

answer 2 : option "d"

answer 3 : option "b"

answer 4 : option "d"

answer 5 : option "b"

answer 6 : option "a"

answer 7 : option "c"

answer 8 : option "d"

answer 9 : option "c"

answer 10 : option "d"

answer 11 : option "c"

answer 12 option "b"

answer 13 option "c"

answer 14 option "c"

answer 15 option "d"

answer 16 option "a"

answer 17 option "b"

answer 18 option "c"

answer 19 option "a"

answer 20 option "a"

answer 21 option "c"

d