Question

The rate of monthly salary of an office assistant increased annually in A.P if he was drawing ₹200 a month during the 11th year and ₹380 a month during 29th year find out his initial salary and the rate of annual increment. Find also his salary at the time of completion of 32 years of service.

Answer 1

Answer:

$\bigstar{\bold{Rate\:of\:increment=Rs\:10\:per\:year}}$

$\bigstar{\bold{Initial\:salary=Rs.100}}$

$\bigstar{\bold{Salary\:at\:32nd\:year=Rs.410}}$

Step-by-step explanation:

$\Large{\underline{\underline{\sf{Given:}}}}$

• During 11th year salary (a₁₁) = Rs. 200 a month
• During 29th year salary (a₂₉) = Rs.380 a month

$\Large{\underline{\underline{\sf{To\:Find:}}}}$

• Initial salary (a₁)
• Rate of annual increment (d)
• Salary at 32nd year (a₃₂)

$\Large{\underline{\underline{\sf{Solution:}}}}$

Rate of annual increment:

➣ Here the rate of annual increment is the common difference of the A.P

➣ The common difference d is given by equation,

    $d=\dfrac{a_m-a_n}{m-n}$

   where $a_m$ = 380, $a_n$ = 200, m = 29, n = 11

➣ Substituting the datas we get.

    $d=\dfrac{380-200}{29-11}$

    $d=\dfrac{180}{18}$

    d = 10

➣ Hence the rate of annual increment is Rs.10 per year

$\boxed{\bold{Rate\:of\:increment=Rs\:10\:per\:year}}$

Initial salary:

➣ Here initial salary is the first term of the A.P

➣ We know that

     a₁₁ = a₁ + 10 d

     a₁ = a₁₁ - 10 d

➣ Substituting the datas,

     a₁ = 200 - 10 × 10

     a₁ = 200 - 100

     a₁ = 100

➣ Hence the initial salary was Rs.100

$\boxed{\bold{Initial\:salary=Rs.100}}$

Salary at 32nd year:

➣ The salary at 32nd year is 32nd term of the A.P

➣ The 32nd term of an A.P is given by

    a₃₂ = a₁ + 31 d

➣ Substitute the datas,

    a₃₂ = 100 + 31 × 10

    a₃₂ = 100 + 310

    a₃₂ = 410

➣ Hence the salary at 32nd year is Rs.410

$\boxed{\bold{Salary\:at\:32nd\:year=Rs.410}}$