Question

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?

Answer 1

Answer:  11th year

   

Step-by-step explanation:

Starting year=1995

Starting Salary=Rs.5000

Increment each year=Rs.200

Final Salary=Rs. 7000

a=5000     d= 200     An=7000

An=a+(n-1)d

7000=5000+(n-1)200

(n-1)200= 2000

n-1=2000/200

n-1=10

n=11

In 11th year his salary will reach  Rs.7000

Answer 2

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 :

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

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

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

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

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