Let us revisit the sequence $t_{n}$ of triangular numbers 1, 3, 6, 10, 15, $\dots$ shown in the given figure. Note that the $n$ th term of this sequence is the sum of the first $n$ natural numbers. Thus, $t_{n} = \frac{n(n + 1)}{2}$ .

Can you use this to find the $10^{\text{th}}$ , $17^{\text{th}}$ and $80^{\text{th}}$ triangular numbers? [Page No. 185]

Answer: Verified

Yes. Each triangular number is the sum of the first $n$ natural numbers, so $t_n = \frac{n(n+1)}{2}$ .

$t_{10} = \frac{10 \times 11}{2} = \frac{110}{2} = 55$

$t_{17} = \frac{17 \times 18}{2} = \frac{306}{2} = 153$

$t_{80} = \frac{80 \times 81}{2} = \frac{6480}{2} = 3240$

(All Q's of this Chapter solved)

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