How many terms are there in the sequence 1, 3 | vedclass

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(B) The given sequence is $1, 3, 6, 10, 15, 21, \dots, 5050$.These are triangular numbers,where the $n$-th term is given by the formula $T_n = \frac{n

Vedclass · Accepted Answer

$100$

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(B) The given sequence is $1, 3, 6, 10, 15, 21, \dots, 5050$.These are triangular numbers,where the $n$-th term is given by the formula $T_n = \frac{n(n+1)}{2}$.We are given that the last term $T_n = 5050$.So,$\frac{n(n+1)}{2} = 5050$.$n(n+1) = 10100$.$n^2 + n - 10100 = 0$.Solving this quadratic equation using the quadratic formula $n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:$n = \frac{-1 \pm \sqrt{1^2 - 4(1)(-10100)}}{2(1)} = \frac{-1 \pm \sqrt{1 + 40400}}{2} = \frac{-1 \pm \sqrt{40401}}{2}$.Since $\sqrt{40401} = 201$,we have $n = \frac{-1 + 201}{2} = \frac{200}{2} = 100$.Thus,there are $100$ terms in the sequence.

How many terms are there in the sequence $1, 3, 6, 10, 15, 21, \dots, 5050$?

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