 Anna has Rs 25,000 in her savings account. Each month, she withdraws Rs 350. How much money is she left with at the end of 41 months? So, in the first month, Anna has Rs 25,000 in her bank account and then she starts taking out Rs 350 every month. So, after the first month, when she takes out Rs 350 from her account, in the second month, she would be having Rs 24,650 remaining in her account. And then again, after the second month, when she again takes out Rs 350, then in the third month, she would be having Rs 24,300 remaining in her account. And she keeps on withdrawing Rs 350 every month, I suppose till she has no money left in her account. And what we need to figure out is, how much money will she have in her account by the end of 41 months? Now, one way to go about it is, we can keep on subtracting Rs 350 from each month's remaining amount for 41 times and we'll eventually get there. But imagine, what if we had to figure out how much money would Anna have by the end of 500th month? I'm assuming she would have none. But even then, this seems like a little bit of hassle. So, let's come up with some other way, using which we can figure out how much this amount is going to be without all this hassle. Well, this is a sequence of numbers and I see a pattern over here. In this sequence, every successive term is 350 less than its preceding term. So, whenever we have a pattern that involves adding or subtracting a constant number to each term to get to the next term, we have an arithmetic progressions. So, all the figures in Anna's saving accounts are in arithmetic progression. And to figure out any term of an arithmetic progression, all we need is the first term, which is 25,000 in this case. This is our first term. And this constant number known as the common difference, which is the difference between two consecutive terms of an AP. So, using these two things, we could figure out any term of this AP and we need to precisely figure out the 41th term of this AP. So now, as you can see, to get to our second term from our first term, we added this common difference once to our first term. Similarly, to get to our third term, we had to add this common difference twice. So, by adding this common difference twice to our first term, we get the third term. So, I'm sure you're observing the pattern here. To get to any term in an AP, let's say to get to an, the n-th term of an AP, all we need to do is, we need to add the first term to n-1 times of the common difference and this would give us any term in an AP. So, we can arrive at this general expression by thinking about terms of an AP. And you don't really need to memorize it, but it really comes in handy when we are solving problems such as these. And if you want to learn more about what an AP is and if you want to look into some more examples, you should watch our previous videos. And with this out of the way, I would want you to pause the video here and figure out the 41th term all by yourself. So, let's solve this. We need to figure out the 41th term. The first term is 25,000. So, I'll write 25,000 instead of a-1. And the term we need to figure out is the 41th one. So, 41-1 is n-1 times d is negative 350. So, I'll write negative 350 in that place. So, let's simplify this to figure out how much money Anna has remaining in her account. So, this would simplify to 25,000 plus 41-1 is 40 times negative 350. So, 40 times 300 is 12,000. And 40 times 50 is 2,000. So, 12,000 and 2,000 is 14,000. So, this simplifies to 25,000 minus 12, sorry, 14,000. And how much is this? This comes out to be 11,000. So, Anna has 11,000 rupees remaining in her bank account by the end of 41 months.