Question

A person has monthly income of rs.1000 and his income is increased by rs.50 of the previous month income.on which term his total salary will exceed rs.100,000?

Answer 1

Answer:

$1982^{th}$ term

Step-by-step explanation:

Monthly income of the person is

1000 , 1050 , 1100 , 1150 . . . . .

This forms an AP where a = 1000 and d = 50

$A_{n}$ = 100000 = a + [(n - 1) d]

⇒ 100000 = 1000 + [(n - 1) x 50]

⇒ 100000 = 1000 + 50n - 50

⇒ 99050 = 50n

⇒ n = 1981

The person's salary equals 100000 on the $1981^{th}$ term

Hence, it exceeds 100000 on the $1982^{th}$ term

Answer 2

Given: monthly income = rs.1000 which is called a, income is increased = rs.50 which is called d (common difference) and his total salary will exceed = rs.100,000

To find: to find term

Step-by-step explanation:

In mathematics arithmetic progression/series is one branch of algebra the consecutive term is the same which is called common difference d, for all term so it called an A.P

According to question

1000,1050,1100..............100000

then,

$an=a+(n-1)d$

$100000=1000+(n-1)50$

$100000=1000+50n -50$

$100000=950+50n$

$99050=50n$

$n=\frac{99050}{50}$

= 1981