Question

1. Subba Rao started work in 1995 at an annual salary of Rs 5000 and received an increment of Rs
200 each year. In which year did his income reach Rs 7000?
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Answer 1

Answer:

2005

Step-by-step explanation:

a = 5000

n = ?

d = 200

tn = a + (n - 1)d

7000 = 5000 + (n - 1) 200

2000 = 200n - 200

2200 = 200n

n = 11

therefore

1995 + 11 - 1

year = 2005

Answer 2

Given :-

• First term (a) = 5000

• common difference (d) = 200

• nth term (an) = 7000

To find :-

• In which year did his income reach Rs 7000 (n) = ?

Solution :-

$\orange{\bigstar}\:\:{\underline{\boxed{\bf\red{a_n=a+(n-1)d}}}}$

$\rm\longrightarrow\:7000=5000+(n-1)\times{200}$

$\rm\longrightarrow\:7000-5000=(n-1)\times{200}$

$\rm\longrightarrow\:2000=(n-1)\times{200}$

$\rm\longrightarrow\dfrac{2000}{200}=n-1$

$\rm\longrightarrow\:10=n-1$

$\rm\longrightarrow\:10+1=n$

$\rm\longrightarrow\:11=n$

$\rm\longrightarrow\:n=11$

Hence,in 11th year his income reach Rs 7000.

More Information :-

Arithmetic progression :- A sequence in which each term differs from its preceding term by a constant contact is called an arithmetic progression which is denoted by AP.

Related to arithmetic mean :-

➝ If a,A,b are in AP we can say that A is the arithmetic mean between a and b and it is abbreviated of AM.