A person saved Rs. 16,500 in all during 10 ye | vedclass

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(A) Let the savings of the first year be $a$. Since the savings increase by $Rs. 100$ every year,this forms an Arithmetic Progression $(AP)$ where the

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$1200$

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(A) Let the savings of the first year be $a$. Since the savings increase by $Rs. 100$ every year,this forms an Arithmetic Progression $(AP)$ where the common difference $d = 100$. The number of years is $n = 10$,and the total savings $S_{10} = 16,500$. The formula for the sum of $n$ terms of an $AP$ is $S_n = \frac{n}{2} [2a + (n - 1)d]$. Substituting the values: $16,500 = \frac{10}{2} [2a + (10 - 1) 	imes 100]$. $16,500 = 5 [2a + 900]$. $3,300 = 2a + 900$. $2a = 3,300 - 900 = 2,400$. $a = 1,200$. Thus,the savings of the first year is $Rs. 1,200$.

$A$ person saved $Rs. 16,500$ in all during $10$ years. After the first year,every year he saved $Rs. 100$ more than what he had saved the previous year. Find his savings of the first year.

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