Question

The 5th term of an A.P is 26 and it's 10th term is 51.find the A.P

Answer 1

Given:

• The 5th term of an AP = 26
• The 10th term of an AP = 51

To Find:

• The AP.

Solution:

It is given that,

$a_5$  =  a+4d =  26  → {equation 1}

$a_{10}$ = a+9d = 51 → {equation 2}

Subtract equations 1 and 2. We get,

⇒ -5d = -25

⇒ d = 5

Now substitute the value of d in equation 1. We get,

⇒ a+4(5) = 26

⇒ a+20 = 26

⇒ a = 26-20 = 6

The general form of an AP is $t_n = (a+(n-1)d$  for this equation,  substitute the values of "a" and "d" for each term and we get the following arithmetic progression.

The first term of AP is given by,

⇒ $t_1$ = 6+(1-1)5 = 6

The second term of the AP is given by,

⇒ $t_2$ = 6+(2-1)5 = 6+5 = 11

The third term of the AP is given by,

⇒ $t_3$ = 6+(3-1)5 = 6+10 = 16

∴ The required AP is 6, 11, 16, and so on