Question

Neera saves Rs. 1,600/- during the first year, Rs. 2,100/- in the second year, Rs. 2,600/- in
the third year. If she continues her saving in this pattern, in how many years will she save
Rs, 38,500/-
(10 year]​

Answer 1

Answer:

10 years.

Step-by-step explanation:

We know that the sum of an A.P. series with a as the first term, d as the common difference and n as the number of terms, is given by

$S_{n}=\frac{n}{2}[2a+(n-1)d]$ ......... (1)

Now, given that Neera saves 1600 Rs. in the first year, 2100 Rs. in the second year, 2600 Rs. in the third year and continued saving in this pattern.

Clearly, she saves in an A.P. series pattern, where the first value is 1600 and the next savings are 500 more than the previous one.

Let us assume that she will save 38500 Rs. in total after n years of savings, then comparing with equation (1), we get

$\frac{n}{2}[2*1600+(n-1)*500]=38500$

⇒1600n+250n²-250n=38500

⇒250n²+1350n-38500=0

⇒5n²+27n-770=0

⇒ n = $\frac{-27+\sqrt{27^{2}-4*5*(-770) } }{2*5}$ {Neglecting the negative root. as n can not be negative}

⇒ n = $\frac{-27+127}{10}$

⇒ n = 10 years.

So, in 10 years Neera will save 38500 Rs. (Answer)