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A Man Accepts a Position with an Initial Salary of ₹5200 per Month. It is Understood that He Will Receive an Automatic Increase of ₹320 in the Very Next Month and Each Month Thereafter. (I) Fi

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Question

A man accepts a position with an initial salary of ₹5200 per month. It is understood that he will receive an automatic increase of ₹320 in the very next month and each month thereafter.
(i) Find his salary for the tenth month.
(ii) What is his total earnings during the first year?

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Solution

We have,
the initial salary, a1 = ₹5200,
the salary of the second month, a2 = ₹5200 + ₹320 = ₹5520,
the salary of the third month, a3 = ₹5520 + ₹320 = ₹5840,

\[\text { As, } a_2 - a_1 = 5520 - 5200 = 320 \text { and } a_3 - a_2 = 5840 - 5520 = 320\]

\[i . e . a_2 - a_1 = a_3 - a_2 \]

\[\text { So, } a_1 , a_2 , a_3 , . . . \text { are in A . P } . \]

\[\text { Also, } a = 5200, d = 320\]

\[\left( i \right) a_{10} = a + \left( 10 - 1 \right)d\]

\[ = 5200 + 9 \times 320\]

\[ = 5200 + 2880\]

\[ = 8080\]

\[\text { So, the salary of the man for the tenth month is } ₹ 8, 080 . \]

\[\left( ii \right) S_{12} = \frac{12}{2}\left[ 2a + \left( 12 - 1 \right)d \right]\]

\[ = 6\left( 2 \times 5200 + 11 \times 320 \right)\]

\[ = 6\left( 10400 + 3520 \right)\]

\[ = 6 \times 13920\]

\[ = 83520\]

\[\text { So, the total earnings of the man during the first year is } ₹ 83, 520 .\]

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Chapter 19: Arithmetic Progression - Exercise 19.7 [Page 50]

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R.D. Sharma Mathematics [English] Class 11
Chapter 19 Arithmetic Progression
Exercise 19.7 | Q 15 | Page 50

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