1. **Find the 10th term of the arithmetic sequence where $a_1=5$ and $d=3$.** The $n$th term of an arithmetic sequence is given by: $$a_n = a_1 + (n-1)d$$ For $n=10$: $$a_{10} = 5 + (10-1)\times 3 = 5 + 9\times 3 = 5 + 27 = 32$$ 2. **Calculate the sum of the first 20 terms when $a_1=2$ and $d=4$.** Sum of first $n$ terms: $$S_n = \frac{n}{2}[2a_1 + (n-1)d]$$ For $n=20$: $$S_{20} = \frac{20}{2}[2\times 2 + (20-1)\times 4] = 10[4 + 76] = 10 \times 80 = 800$$ 3. **If the 5th term is 18 and $d=3$, find $a_1$.** Use the formula: $$a_5 = a_1 + 4d$$ $$18 = a_1 + 4\times 3$$ $$a_1 = 18 - 12 = 6$$ 4. **Write the first 5 terms with $a_1=7$ and $d=-1$.** Terms: $$a_1=7$$ $$a_2=7-1=6$$ $$a_3=7-2=5$$ $$a_4=7-3=4$$ $$a_5=7-4=3$$ Sequence: 7,6,5,4,3 5. **Sum of the series $4 + 7 + 10 + ...$ up to 15 terms.** $a_1=4, d=3, n=15$ $$S_{15} = \frac{15}{2}[2\times 4 + 14 \times 3] = \frac{15}{2}[8 + 42] = \frac{15}{2} \times 50 = 15 \times 25 = 375$$ 6. **Sum of first 8 terms is 124 and $a_1=6$, find $d$.** $$S_8 = \frac{8}{2}[2\times 6 + 7d] = 124$$ $$4(12 + 7d) = 124$$ $$48 + 28d = 124$$ $$28d = 76$$ $$d = \frac{76}{28} = \frac{19}{7} = 2.7143$$ 7. **Given $a_1=12$ and $S_{15} = 255$, find $d$.** $$255 = \frac{15}{2}[2\times 12 + 14d]$$ $$255 = \frac{15}{2}(24 + 14d)$$ $$255 = 7.5(24 + 14d)$$ $$24 + 14d = \frac{255}{7.5} = 34$$ $$14d = 10$$ $$d = \frac{10}{14} = \frac{5}{7} = 0.7143$$ 8. **Find the 25th term with $a_1=3$ and $S_{25}=975$.** Sum formula: $$S_{25} = \frac{25}{2}[2\times 3 + 24d] = 975$$ $$\frac{25}{2}(6 + 24d) = 975$$ $$12.5(6 + 24d) = 975$$ $$6 + 24d = \frac{975}{12.5} = 78$$ $$24d = 72$$ $$d = 3$$ 25th term: $$a_{25} = 3 + 24\times 3 = 3 + 72 = 75$$ 9. **Given 3rd term is 20 and 7th term is 40, find $a_1$ and $d$.** From the formula: $$a_3 = a_1 + 2d = 20$$ $$a_7 = a_1 + 6d = 40$$ Subtract equations: $$(a_1 + 6d) - (a_1 + 2d) = 40 - 20$$ $$4d = 20$$ $$d = 5$$ Plug back in: $$a_1 + 2\times 5 = 20$$ $$a_1 = 20 - 10 = 10$$ 10. **Sum of first $n$ terms: $S_n=3n^2+2n$. Find $a_1$ and $d$.** Recall: $$a_1 = S_1$$ Calculate: $$S_1 = 3(1)^2 + 2(1) = 3 + 2 = 5$$ So, $a_1=5$ Also, 1st term: $$a_1=5$$ Find $a_2$: $$a_2 = S_2 - S_1 = [3(2)^2 + 2(2)] - 5 = (12 +4) - 5 = 16 - 5 = 11$$ $d = a_2 - a_1 = 11 - 5 = 6$ 11. **Sum of 750 for first $n$ terms, $a_1=5$, $d=4$. Find $n$.** Sum formula: $$750 = \frac{n}{2}[2\times 5 + (n-1)4] = \frac{n}{2}[10 + 4n - 4] = \frac{n}{2}(4n +6)$$ Multiply both sides: $$750 \times 2 = n(4n + 6)$$ $$1500 = 4n^2 + 6n$$ Rearranged: $$4n^2 + 6n - 1500 = 0$$ Divide all by 2: $$2n^2 + 3n - 750 = 0$$ Use quadratic formula: $$n = \frac{-3 \pm \sqrt{3^2 - 4 \times 2 \times (-750)}}{2\times 2} = \frac{-3 \pm \sqrt{9 + 6000}}{4} = \frac{-3 \pm \sqrt{6009}}{4}$$ Approximate: $$\sqrt{6009} \approx 77.5$$ Take positive root: $$n = \frac{-3 + 77.5}{4} = \frac{74.5}{4} = 18.625$$ Since $n$ must be an integer, $n=19$ (sum will exceed 750 at $n=19$) 12. **Prove average of first and $n$th term equals average of all $n$ terms.** Average of first and $n$th term: $$\frac{a_1 + a_n}{2}$$ Sum of $n$ terms: $$S_n = \frac{n}{2}(a_1 + a_n)$$ Average of all $n$ terms: $$\frac{S_n}{n} = \frac{\frac{n}{2}(a_1 + a_n)}{n} = \frac{a_1 + a_n}{2}$$ This proves the averages are equal. 13. **Sum = 210 and sum of squares of first 5 terms = 979. Find $a_1$ and $d$.** Sum formula: $$S_5 = \frac{5}{2}[2a_1 + 4d] = 210$$ $$5[ a_1 + 2d ] = 210$$ $$a_1 + 2d = 42$$ Terms: $$a_k = a_1 + (k-1)d$$ Sum of squares: $$\sum_{k=1}^5 (a_1 + (k-1)d)^2 = 979$$ Expand sum of squares: $$\sum_{k=0}^4 (a_1 + kd)^2 = \sum (a_1^2 + 2a_1kd + k^2 d^2) = 5a_1^2 + 2a_1 d \sum k + d^2 \sum k^2$$ Calculate sums: $$\sum k = 0+1+2+3+4 = 10$$ $$\sum k^2 = 0^2+1+4+9+16 = 30$$ So: $$5a_1^2 + 2a_1 d \times 10 + d^2 \times 30 = 979$$ $$5a_1^2 + 20 a_1 d + 30 d^2 = 979$$ From $a_1 + 2d = 42$, we get: $$a_1 = 42 - 2d$$ Plug into sum of squares: $$5(42-2d)^2 + 20(42-2d)d + 30 d^2 = 979$$ Expand: $$5(1764 -168 d + 4 d^2) + 840 d - 40 d^2 + 30 d^2 = 979$$ $$5\times 1764 -840 d + 20 d^2 + 840 d - 40 d^2 + 30 d^2 = 979$$ $$8820 + ( -840 d + 840 d ) + (20 - 40 + 30) d^2 = 979$$ $$8820 + 10 d^2 = 979$$ Simplify: $$10 d^2 = 979 - 8820 = -7841$$ Since $d^2$ can't be negative, check work for arithmetic or calculation errors. Recalculate sums: Sum of $k^2$ is correct: 0+1+4+9+16 = 30 Sum of $k$ is 10 Double check substitution: $$5(42-2d)^2 = 5(1764 - 168 d + 4 d^2) = 8820 - 840 d + 20 d^2$$ $$20 a_1 d = 20 (42 - 2d) d = 840 d - 40 d^2$$ Sum: $$8820 - 840 d + 20 d^2 + 840 d - 40 d^2 + 30 d^2 = 8820 + ( - 840 d +840 d ) + (20 -40 +30) d^2 = 8820 + 10 d^2$$ So all correct. The equation is: $$8820 + 10 d^2 = 979$$ This leads to contradiction since 979 < 8820. Thus no real solution. The problem might have a typo or no valid arithmetic sequence with those sums. 14. **Sum of all integers between 100 and 500 divisible by 7.** Find smallest integer ≥100 divisible by 7: $$\lceil \frac{100}{7} \rceil = 15$$ $$15 \times 7 = 105$$ Find largest integer ≤500 divisible by 7: $$\lfloor \frac{500}{7} \rfloor = 71$$ $$71 \times 7 = 497$$ Number of terms: $$n = 71 - 15 +1 = 57$$ Sum: $$S_n = \frac{n}{2}(a_1 + a_n) = \frac{57}{2}(105 + 497) = 28.5 \times 602 = 17157$$ 15. **Derive formula for sum of first $n$ odd numbers using arithmetic series.** Odd numbers sequence: $$1, 3, 5, ..., (2n-1)$$ $a_1 = 1, d=2$ Sum: $$S_n = \frac{n}{2}[2 \times 1 + (n-1)2] = \frac{n}{2}[2 + 2n - 2] = \frac{n}{2}(2n) = n^2$$ **Hence, the sum of the first $n$ odd numbers is $n^2$.**