Yasmeen saves Rs 32, Rs 36, Rs 40 in months 1,2,3 and continues so (AP with \(d=4\)). In how many months will she save Rs 2000?
25 months
Step 1: The savings form an Arithmetic Progression (AP): 32, 36, 40, ...
Here, first term \(a = 32\), common difference \(d = 4\).
Step 2: To find in how many months Yasmeen saves Rs 2000, we need the formula for the sum of \(n\) terms of an AP:
\[ S_n = \dfrac{n}{2} \big[2a + (n-1)d \big] \]
Step 3: Substitute the values of \(a=32\) and \(d=4\):
\[ S_n = \dfrac{n}{2} [2 \times 32 + (n-1) \times 4] \]
\[ S_n = \dfrac{n}{2} [64 + 4n - 4] \]
\[ S_n = \dfrac{n}{2} [4n + 60] \]
\[ S_n = n(2n + 30) \]
Step 4: We want \(S_n = 2000\):
\[ n(2n + 30) = 2000 \]
Step 5: Expand and simplify:
\[ 2n^2 + 30n = 2000 \]
\[ 2n^2 + 30n - 2000 = 0 \]
Divide the whole equation by 2:
\[ n^2 + 15n - 1000 = 0 \]
Step 6: Solve the quadratic equation \(n^2 + 15n - 1000 = 0\):
\[ n = \dfrac{-15 \pm \sqrt{15^2 - 4(1)(-1000)}}{2} \]
\[ n = \dfrac{-15 \pm \sqrt{225 + 4000}}{2} \]
\[ n = \dfrac{-15 \pm \sqrt{4225}}{2} \]
\[ n = \dfrac{-15 \pm 65}{2} \]
Step 7: Calculate the two roots:
\[ n = \dfrac{-15 + 65}{2} = \dfrac{50}{2} = 25 \]
\[ n = \dfrac{-15 - 65}{2} = -40 \; (\text{not possible since months cannot be negative}) \]
Final Answer: Yasmeen will save Rs 2000 in 25 months.