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


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 \(\rm{Rs}\, 20.75\)


\({a_n} = a + \left( {n - 1} \right)d\) 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.


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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