Question

subba rao started work in 1995 at an annual salary of rs 5000 and recived an increment of rs 200 each year in which year did his income reach rs 7000

Answer 1

The annual salary received by Subba Rao in the years 1995,1996,1997..... is 5000, 5200, 5400,...... 7000 

 The incomes that Subba Rao obtained in various years are in A.P. as every year, his salary is increased by Rs 200.


Therefore, the salaries of each year after 1995 are
5000, 5200, 5400, …

Given:

a = 5000
d = 200


Let after nth year, his salary be Rs 7000.
 an = a + (n − 1) d

7000 = 5000 + (n − 1) 200
200(n − 1) = 2000
(n − 1) = 10
n = 11


Thus in 11th year, his salary will be Rs 7000.

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Answer 2

Step-by-step explanation:

AnswEr

• Subba started work in 1995 at annual salary of Rs 5000 & recieved an increment of Rs 200 each year.

• We've to find out in which year his income reach at Rs 7000.

⠀⠀${\underline{\sf{\bigstar\: According \ to \ Question \: Now :}}}$

Salary in 1996 [5000 + 200] = 5200

Salary in 1997 [5200 + 200] = 5400

Salary in 1998 [5400 + 200] = 5600

5200, 5400, 5600... so on.

$\star\:\boxed{\textsf{This is in Arithmetic Progression}}$

For any Arithmetic Progression ( AP ), the nth term Formula is Given by :

$\star\: \boxed{\sf{\pink{a_{n} = a + (n - 1)d}}}$

$\bf{Here}\begin{cases}\sf{ \: a_{n} = 7000}\\\sf{\: First \ term \ (a) = 5000}\\\sf{ \: Common \ difference \ (d) = 200}\end{cases}$

$\underline{\bf{\dag} \:\mathfrak{Substituting \ Values \ in \ the \ formula \ :}}$

$:\implies\sf 7000 = 5000 + (n - 1) 200 \\\\\\:\implies\sf 7000 - 5000 = (n -1) 200 \\\\\\:\implies\sf 2000 = (n - 1) 200\\\\\\:\implies\sf n - 1 = \cancel\dfrac{2000}{200}\\\\\\:\implies\sf n - 1 = 10 \\\\\\:\implies\sf n = 10 + 1\\\\\\:\implies\boxed{\frak{\purple{n = 11}}}$

$\therefore\underline{\textsf{ Hence, in 11th years subba's salary will reach at \textbf{Rs \: 7000}}}.$⠀⠀⠀⠀