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in Arithmetic Progression by (50.9k points)
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A man saves ₹4000 during the first year, ₹5000 during the second year and in this way he increases his savings by ₹1000 every year. Find in what time his savings will be ₹85000.

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A Man saves some amount of money every year.

In the first year, he saves Rs.4000.

In the next year, he saves Rs.5000.

Like this, he increases his savings by Rs.1000 ever year.

Given a total amount of Rs. 85000 is saved in some ‘n’ years.

According to the above information the savings in every year are in Arithmetic Progression.

First year savings = a = Rs.4000

Increase in every year savings = d = Rs.1000

Total savings (sn) = Rs.85000

n2 + 7×n - 170 = 0

(n + 17) ×(n - 10) = 0

n = - 17 or n = 10

Since the number of years cannot be negative, n = 10.

After 10 years his savings will become Rs.85000.

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