Question

In an AP. the third term is four times the first term, and the sixth term is 17; find the series.

Answer 1

let a be the first term and D be the common difference
then a3=4a
a+2d =4a
a-4a+2d=0
-3a+2d=0
3a-2d=0..........(i)

and a6=17
a+5d=17...........(ii)

solving (i) &(ii), we get
a=2 & d=3
therefote ap is
2,5,8,11,.......

Answer 2

Answer:

The series is 2,5,8,...

Step-by-step explanation:

Given : In an AP. the third term is four times the first term, and the sixth term is 17.

To find : The series ?

Solution :

The Arithmetic series is $a,a+d,a+2d,a+3d,..$

The first term is $a_1=a$

The third term is $a_3=a+2d$

The third term is four times the first term,

i.e. $a+2d=4a$

$3a=2d$

$a=\frac{2d}{3}$ ....(1)

The sixth term is $a_6=a+5d$

The sixth term is 17.

i.e. $a+5d=17$

Substitute a from (1),

$\frac{2d}{3}+5d=17$

$\frac{2d+15d}{3}=17$

$\frac{17d}{3}=17$

$d=3$

Substitute in (1),

$a=\frac{2(3)}{3}=2$

So, a=2 and d=3

The series is 2,5,8,...