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Home » A man saved ₹ 16500 in ten years. In each year after the first he saved ₹ 100/- more than he did in the preceding year. How much did he saved in the first year?
A man saved ₹ 16500 in ten years. In each year after the first he saved ₹ 100/- more than he did in the preceding year. How much did he saved in the first year?
Arithmetic Progressions | CBSE | Class 11 | Exercise 19.7 | Maths | RD Sharma
A man saved ₹ 16500 in ten years. In each year after the first he saved ₹ 100/- more than he did in the preceding year. How much did he saved in the first year?

Solution:

As per the question:
A man saved ₹16500 in ten years

Let be his savings in the first year be ₹ x

Every year his savings increased by ₹ 100.

Therefore,

A.P will be x, 100 + x, 200 + x \dots \dots.

In which, x is first term and

Common difference, d = 100 + x - x = 100

It is known that, S_n is the sum of n terms of an A.P

Using the formula,

S_{n}=n / 2[2 a+(n-1) d]

Where, the first term is a, the common difference is d and n is number of terms in an A.P.
Given that,
\begin{array}{l} S_{n}=16500 \text { and } n=10 \\ S_{10}=10 / 2[2 x+(10-1) 100] \\ 16500=5\{2 x+9(100)\} \\ 16500=5(2 x+900) \\ 16500=10 x+4500 \\ -10 x=4500-16500 \\ -10 x=-12000 \\ x=12000 / 10 \\ =1200 \end{array}
As a result, his saving in the first year is ₹ 1200 .

i

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