1. ## I want check my answer

Q 1 : If the second term = 6 and fifteenth = 25 of sequence find the first term ?

2. Originally Posted by r-soy

Q 1 : If the second term = 6 and fifteenth = 25 of sequence find the first term ?
$a_n = a_1 + (n-1)d$

$a_2 = a_1 + d$

$6 = a_1 + d$

$a_{15} = a_1 + 14d$

$25 = a_1 + 14d$

$19 = 13d$

$d = \frac{19}{13}$

$a_1 = 6 - d = 6 - \frac{19}{13} = \frac{59}{13}$

3. Originally Posted by r-soy

Q 1 : If the second term = 6 and fifteenth = 25 of sequence find the first term ?
No, because you haven't answered the question fully. You've been asked to find the first term a but you found the common difference d.

Your working seems fine though. Simply sub your value of d into one of your original equations to find a.

$U_2 = a + d = 6$ (eq1)

$U_{15} = a + 14d = 25$ (eq2)

Rearrange eq1:

$d = 6-a$ (eq3)

sub eq3 into eq2

$a+14(6-a) = 25$

$-13a = -59$

$a = \frac{59}{13}$

4. thanks

5. Originally Posted by r-soy
thanks, but why we d write

$
U_2 = a + 5d = 6
$
Neither me nor skeeter said that? You don't write 5d because there is only d in the second term.

It arises from the definition of an arithmetic sequence