# I want check my answer

• Jan 31st 2010, 09:58 AM
r-soy

Q 1 : If the second term = 6 and fifteenth = 25 of sequence find the first term ?
• Jan 31st 2010, 10:06 AM
skeeter
Quote:

Originally Posted by r-soy

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

$a_n = a_1 + (n-1)d$

$a_2 = a_1 + d$

$6 = a_1 + d$

$a_{15} = a_1 + 14d$

$25 = a_1 + 14d$

$19 = 13d$

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

$a_1 = 6 - d = 6 - \frac{19}{13} = \frac{59}{13}$
• Jan 31st 2010, 10:08 AM
e^(i*pi)
Quote:

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

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

$-13a = -59$

$a = \frac{59}{13}$
• Jan 31st 2010, 10:59 AM
r-soy
thanks
• Jan 31st 2010, 11:07 AM
e^(i*pi)
Quote:

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