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A Man Saves Rs 32 During the First Year. Rs 36 in the Second Year and in this Way He Increases His Savings by Rs 4 Every Year. Find in What Time His Saving Will Be Rs 200. - Mathematics

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Question

A man saves Rs 32 during the first year. Rs 36 in the second year and in this way he increases his savings by Rs 4 every year. Find in what time his saving will be Rs 200.

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Solution

Let n be the time in which the man saved Rs 200.
Here, d= 4, a = 32
We know:

\[S_n = \frac{n}{2}\left\{ 2a + \left( n - 1 \right)d \right\}\]

\[ \Rightarrow 200 = \frac{n}{2}\left\{ 2 \times 32 + \left( n - 1 \right)4 \right\}\]

\[ \Rightarrow 400 = 64n + 4 n^2 - 4n\]

\[ \Rightarrow 4 n^2 + 60n - 400 = 0\]

\[ \Rightarrow n^2 + 15n - 100 = 0\]

\[ \Rightarrow n^2 + 20n - 5n - 100 = 0\]

\[ \Rightarrow \left( n + 20 \right)\left( n - 5 \right) = 0\]

\[ \Rightarrow n = 5, n = - 20 \left( \text  { Rejecting the negative value } \right)\]

Therefore, the man took 5 years to save Rs 200.

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Chapter 19: Arithmetic Progression - Exercise 19.7 [Page 49]

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RD Sharma Mathematics [English] Class 11
Chapter 19 Arithmetic Progression
Exercise 19.7 | Q 2 | Page 49

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