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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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: 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
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.
⠀⠀[tex]{\underline{\sf{\bigstar\: According \ to \ Question \: Now :}}}\\ \\[/tex]
5200, 5400, 5600... so on.
[tex]\star\:\boxed{\textsf{This is in Arithmetic Progression}}[/tex]
[tex]\\[/tex]
For any Arithmetic Progression ( AP ), the nth term Formula is Given by :
[tex]\star\: \boxed{\sf{\pink{a_{n} = a + (n - 1)d}}}[/tex]
[tex]\bf{Here}\begin{cases}\sf{ \: a_{n} = 7000}\\\sf{\: First \ term \ (a) = 5000}\\\sf{ \: Common \ difference \ (d) = 200}\end{cases}[/tex]
[tex]\underline{\bf{\dag} \:\mathfrak{Substituting \ Values \ in \ the \ formula \ :}}[/tex]
[tex]:\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}}}[/tex]
[tex]\therefore\underline{\textsf{ Hence, in 11th years subba's salary will reach at \textbf{Rs \: 7000}}}. \\ [/tex]⠀⠀⠀⠀