Question

The sum of the first six terms of an AP is 345.the difference between the sixth and the first term is 55 find the first six terms.

Answer 1

Let the first term be a
and common difference be d
6th term = a +(6-1)d = a -5d
A/q
(a+5d) -a = 55
⇒5d = 55
⇒d = 11
also, 
S6 = (6/2)[2a+(6-1)×11] = 345
⇒2a +55 = 115
⇒2a = 60
⇒a = 30
so AP = 30,41,52,63,74,85,96.....so on

Answer 2

The sum of six terms is 345.

$a + a + d + a + 3d + a + 3d + a + 4d + a + 5d = 345$
$6a + 15d = 345$

The difference between the last term and the first term is 55.

$(a + 5d) - (a) = 55 \\ \\ 5d = 55 \\ \\ d = 11$

Substitute in 1 to find the first term of the AP.

$6a + 15d = 345 \\ \\ 6a + 15 (11) = 345 \\ \\ 6a + 165 = 345 \\ \\ 6a = 345 - 165 \\ \\ 6a = 180 \\ \\ a = 30$

Now, we know that a = 30 and d = 11.

AP is: 30, 41, 52, 63, 74, 85.