\( \begingroup \)Question 1 The $\var{n}$th term of an arithmetic series is $\simplify{{t1}+({n}-1)*{d}}$ The common difference is $\var{d}$ a) Find the first termExpected answer: Save answer Score: 0/1 Feedback for a). Show feedback.The feedback has changed.This feedback is based on your last submitted answer. Save your changed answer to get updated feedback. Or, you could: ⤺ Go back to the previous part There's nothing more to do from here.b) Find the sum of the first $\var{tsum}$ termsExpected answer: Save answer Score: 0/1 Feedback for b). Show feedback.The feedback has changed.This feedback is based on your last submitted answer. Save your changed answer to get updated feedback. Or, you could: ⤺ Go back to the previous part There's nothing more to do from here. Advice To find the first term, we note that the $\var{n}$th term is $\simplify{{t1}+({n}-1)*{d}}$ and the common difference is $\var{d}$ hence we can calculate the first term as $\simplify{{t1}+({n}-1)*{d}}-(\var{n}-1)\times\var{d}$ To find the sum of the first $\var{tsum}$ terms we used the known formula $S_n = \frac{n(a_1 + a_n)}{2}$ where $a_i$ is the $i$th term of the arithmetic series. Since we know how to calculate $a_n$ from our above considerations, we know that $S_n = \frac{n(a_1+a_1+(n-1)\times d)}{2}$ which simplifies to $S_n = \frac{n(2\times\var{t1}+(n-1)\var{d})}{2}$ \( \endgroup \) Score: 0/2 Total 0/2 Move to the next questionTry another question like this oneReveal answers