A person saves ₹1000 more than he did the previous year. If he saves ₹2000 in the first year, in how many years will he save ₹170000?

Explanation

The savings form an Arithmetic Progression with a=2000, d=1000, and S_n=170000. Using the sum formula S_n = \frac{n}{2}[2a+(n-1)d], we get 170000 = \frac{n}{2}[4000+(n-1)1000]. Solving n^2+3n-340=0 yields n=17.

Related questions on Algebra

• If p + q + r = 0, then what is z^{\frac{p^2}{qr}} \times z^{\frac{q^2}{rp}} \times z^{\frac{r^2}{pq}} equal to ?
• What is the value of k for which (k^2 - 5k + 4)x^2 + (k^2 - 3k - 4)x + (k^2 - 4k) = 0 is an identity ?
• If \frac{a^2}{b^2 + c^2} = \frac{b^2}{c^2 + a^2} = \frac{c^2}{a^2 + b^2}, then what is the value of a^4 + b^4 + c^4 equal to ?
• If \frac{1}{x} = \frac{1}{p} + \frac{1}{q}, then what is \frac{pq}{p^2 - q^2}\left(\frac{x + p}{x - p} - \frac{x + q}{x - q}\right) equa...
• If (x - 5) is the HCF of x^2 - x - p and x^2 - qx - 10, then what is the value of (p + q) ?