Question

The third term of an arithmetic sequence is 15 and when 20 is added to the third term the

Answer 1

Answer:

Common difference = 4

Position of 103 = 25th term

Step-by-step explanation:

a)

According to the given information:

$a_3 + 20 = a_8$

$a+ (3-1)d + 20 = a+ (8-1)d$

$\cancel a+ 2d + 20 = \cancel a + 7d$

$2d + 20 = 7d$

$20 = 7d-2d$

$20 = 5d$

$d = \frac{20}{5}$

$\purple {\bold {d = 4}}$

Common difference = 4

b)

$\because a_3 = 15....(given)$

$\therefore a + (3-1)d = 15$

$\therefore a + 2d = 15$

$\therefore a + 2\times 4= 15$

$\therefore a + 8= 15$

$\therefore a = 15-8$

$\red{\bold {\therefore a = 7}}$

$\because a_n = a + (n - 1) d$

$\therefore 103 = 7 + (n - 1)4$

$\therefore 103 - 7 = (n - 1)4$

$\therefore 96 = (n - 1)4$

$\therefore \frac{96}{4}= n - 1$

$\therefore 24= n - 1$

$\therefore 24+1= n$

$\blue{\bold {\therefore n = 25}}$

So, the position of 103 in this sequence is 25th term.