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 i

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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Eloise 4 days 2021-10-11T08:17:07+00:00 2 Answers 0 views 0

Answers ( )

    0
    2021-10-11T08:18:08+00:00

    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

    0
    2021-10-11T08:18:48+00:00

    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.

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