SERIES DOUBT

Hey! I totally get why this series problem feels confusing at first glance. Sadly there is no fast way to do questions like these, they usually rely on finding the patterns. Here's the solution, let me know if you feel stuck in a particular part (also this question is from Algebra, please post it in the correct farmhouse next time, you're in Arithmetic currently!) Looking at the given terms:

• $t_1 = 1$
• $t_2 = 2 + 3 = 5$
• $t_3 = 4 + 5 + 6 = 15$
• $t_4 = 7 + 8 + 9 + 10 = 34$

Each term is the sum of consecutive integers, and the number of integers being added increases by 1 each time.

• $t_1$: Sum of 1 consecutive integer starting from 1
• $t_2$: Sum of 2 consecutive integers starting from 2
• $t_3$: Sum of 3 consecutive integers starting from 4
• $t_4$: Sum of 4 consecutive integers starting from 7

Notice how the starting numbers are: 1, 2, 4, 7... The differences between consecutive starting numbers are: 1, 2, 3... (increasing by 1 each time)

For $t_{10}$, we need the sum of 10 consecutive integers. But where do we start? Following the pattern:

• Start of $t_5$: 7 + 4 = 11
• Start of $t_6$: 11 + 5 = 16
• Start of $t_7$: 16 + 6 = 22
• Start of $t_8$: 22 + 7 = 29
• Start of $t_9$: 29 + 8 = 37
• Start of $t_{10}$: 37 + 9 = 46

So $t_{10}$ = sum of 10 consecutive integers starting from 46 $t_{10} = 46 + 47 + 48 + 49 + 50 + 51 + 52 + 53 + 54 + 55$ Instead of adding these one by one, use the formula for sum of consecutive integers: Sum = $\frac{n}{2} \times (\text{first term} + \text{last term})$ Where n = 10, first term = 46, last term = 55 $t_{10} = \frac{10}{2} \times (46 + 55) = 5 \times 101 = 505$Answer: 505 (Option B)