Question

the 4th term of an a. p. is equal to 3 times the first term and 7th term exceeds twice the 3rd term by 1. I find 1st term and common diffrence

Answer 1

Therefore,
first term = 3
common difference = 2
Hope this helps u.........if it helps u mark me as brainliest.

Answer 2

↑ Here is your answer ↓
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Given that,

$a_4 = 3a$

$a + 3d = 3a$

By opening,

$-2a + 3d = 0$

$2a - 3d = 0$

$2a = 3d$

$\therefore a = 3d/2 = 1.5d --(1)$

Now,

$a_7 = 2*a_3 + 1$

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

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

Open now,

$a + 6d - 2a - 4d - 1 = 0$

$-a + 2d - 1 = 0 --(2)$

Sub (1) in (2),

$-1.5d + 2d - 1 = 0$

$0.5d - 1 = 0$

$0.5d = 1$

$d = 1/0.5$

$d = 2$

To find 'a',

$a = 1.5d$

$a = 1.5*2$

$a=3$

Thank you so much for asking ^^