Question

Find the number of terms of the A.P.
7, 11, 15, ..., 139?

Answer 1

Answer

34

Explanation

Given A.P. 7, 11, 15.......139

Here, a = 7

d = 11 - 7 = 4

and last term = 139

Let there be n terms in the A.P., then 139 will be the nth term of the A.P.

Apply the formula of general term of an A.P.

$\sf T_n = a + (n - 1)d$

Here,

139 = 7 + (n - 1)4

⇒ 139 - 7 = (n - 1)4

⇒ 132 = (n - 1)4

⇒ n - 1 = 132/4

⇒ n - 1 = 33

⇒ n = 33 + 1 = 34

Hence, there are 34 terms in the A.P.

Answer 2

Answer:

• a = 7
• d = 11 - 7 = 4
• an = 139
• n = ?

We know that, an = a + (n - 1)d

139 = 7 + (n - 1)4

139 = 7 + 4n - 4

139 = 3 + 4n

4n = 139 - 3

4n = 136

n = 136/4

n = 34

$\rule{200}2$

• an = a + (n - 1)d

• Sn = n/2 (2a + (n - 1)d)

• Arithmetic progression is a series of numbers with some difference in between the terms which is known as common difference.

• a stands for the first term.

• d stands for the common difference between the terms.

• an or called the nth term, stands for any term of the arithmetic progression Eg:- a2, which stands for the second term of the arithmetic progression.

• n stands for the number of terms.

• Sn stands for the sum of terms.

$\rule{200}2$