Question

In an Arithmetic Progression sixth term is one more than twice the third
term. The sum of the fourth and fifth terms is five times the second term
Find the tenth term of the Arithmetic Progression
board​

Answer 1

Solution

nth term of AP = $a_n\: =\: a \: +\: (n-1)d$

First term 'a'

Sequence number of term 'n'

Common difference 'd'

Then AP will be a, a+d, a+2d, a+3d...so on

For e.g. $a_n\: =\: a \: +\: (n-1)d$

$a_3\: =\: a \: +\: (3-1)d$

$a_3\: =\: a \: +\: 2d$

As per question

"Sixth term $a_6$ is one more than twice the Third $a_3$ term"

$a_6\: =\: 2(a_3)\: +\: 1$

$a\: + \:5d\: =\: 2(a\: +\: 2d)\: +\: 1$

$a\: + \:5d\: =\: 2a\: +\: 4d\: +\: 1$

$5d\: - \: 4d \:=\: 2a\: - \: a\: +\: 1$

$d\:=\: a\: +\: 1$

The sum of the fourth $a_4$ and fifth $a_5$ terms is five times the second $a_2$ term

$a_4\: +\: a_5 \: =\: 5a_2$

$(a+3d)\: +\: (a+4d) \: =\: 5(a+d)$

$2a \:+\: 7d \: =\: 5a\: +\: 5d$

$7d \:+\: 5d \: =\: 5a\: - \: 2a$

$2(a+1) \: =\: 3a$

$2a\: +\: 2\: =\: 3a$

$3a\: - \: 2a\: =\: 2$

$3a\: - \: 2a\: =\: 2$

$a \: =\: 2$

d = (a + 1) = (2 +1) = 3

$a_1\: =\: 2$

$a_2\: =\: (a\: +\: d)\: =\: (2+3)\: =\: 5$

$a_3\: =\: (a\: +\: 2d)\: =\: (2+6)\: =\: 8$

.

.

.

.

$a_{10}\: =\: (a\: +\: 9d)\: =\: (2+9\times3)\: =\: 29$

AP = 2,5,8,....., 29(10th term)

So 10th term of AP is 29