# Ex.5.2 Q20 Arithmetic Progressions Solution - NCERT Maths Class 10

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## Question

Ramkali saved $$\rm{Rs}\, 5$$ in the first week of a year and then increased her weekly saving by $$\rm{Rs}\, 1.75.$$ If in the $$n^\rm{th}$$ week, her weekly savings become $$\rm{Rs}\, 20.75,$$ find $$n.$$

Video Solution
Arithmetic Progressions
Ex 5.2 | Question 20

## Text Solution

What is Known:?

Savings in first week $$\rm{Rs}\, 5$$ and increment of $$\rm{Rs}\, 1.75$$ weekly in savings.

What is Unknown?

Week in which her savings become $$\rm{Rs}\, 20.75$$

Reasoning:

$${a_n} = a + \left( {n - 1} \right)d$$ is the general term of AP. Where $${a_n}$$ is the $$n^\rm{th}$$ term, $$a$$ is the first term, $$d$$ is the common difference and $$n$$ is the number of terms.

Steps:

From the given data, Ramkali’s savings in the consecutive weeks are

$$\rm{Rs}\, 5, \,\rm{Rs}\, (5+1.75),\, \rm{Rs}\, (5+2×1.75), \\\rm{Rs}\, (5+3×1.75)\, \dots$$ and so on

Hence in $$n^\rm{th}$$ weeks savings, $$\rm{Rs}\, [5+(n−1)×1.75] = \rm{Rs}\, 20.75$$

Now from the above we know that

\begin{align} a&=5 \\ d&=1.75 \\{{a}_{n}}&=20.75 \\n&=?\end{align}

We know that the $$n^\rm{th}$$ term of an A.P. Series,

\begin{align}{a_n} &= a + (n - 1)d\\20.75& = 5 + (n - 1)1.75\\15.75 &= (n - 1)1.75\n - 1)& = \frac{{15.75}}{{1.75}}\\n - 1 &= \frac{{1575}}{{175}}\\n - 1 &= 9\\n &= 10\end{align} The answer is \(n = 10

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