Sequences and Series — Free Algebra 2 Review Games.

Common difference is 3: 11 + 3 = 14.

Each term is multiplied by 2 (common ratio), so it's geometric.

9 - 4 = 5, 14 - 9 = 5. Common difference is 5.

6/2 = 3, 18/6 = 3. Common ratio is 3.

a_n = a_1 + (n-1)d where d is the common difference.

a_10 = 3 + 9(4) = 3 + 36 = 39.

a_5 = 2 * 3^4 = 2 * 81 = 162.

Sum = 1+3+5+7+9 = 25. Or use formula: n^2 = 5^2 = 25.

Sigma notation compactly writes the sum of multiple terms.

S_n = n/2 * (first + last) = n/2 * (a_1 + a_n).

a_20 = 5+19(3) = 62. S_20 = 20/2*(5+62) = 10*67 = 670.

S = a_1/(1-r) = 4/(1-0.5) = 4/0.5 = 8.

The series converges only when the absolute value of the ratio is less than 1.

Each term is 2k where k goes from 1 to 5.

S = a_1(r^n - 1)/(r-1) = 3(2^5 - 1)/(2-1) = 3(31) = 93.

The sequence \(3, 7, 11, 15\) has a constant difference of \(4\) between consecutive terms, which is the defining rule of an arithmetic sequence. The sequence "\(2, 4, 8, 16\)" instead has a constant ratio of \(2\), making it geometric rather than arithmetic. Always test consecutive terms by subtraction first: if the difference is constant, the sequence is arithmetic.

The sequence \(5, 10, 20, 40\) has a constant ratio of \(2\) between consecutive terms, which defines a geometric sequence. The sequence "\(6, 9, 12, 15\)" instead has a constant difference of \(3\), so it is arithmetic, not geometric. Always divide consecutive terms to check for a constant ratio when identifying geometric sequences.

Subtracting consecutive terms gives \(3 - 7 = -4\), so the common difference is \(-4\). The value "\(4\)" is wrong because it reverses the sign, ignoring that the sequence is decreasing. To find the common difference, always subtract a term from the term that follows it, keeping the sign consistent.

Dividing any term by the previous term, \(\frac{50}{100} = \frac{1}{2}\), shows the common ratio is \(\frac{1}{2}\). The value "\(2\)" is incorrect because it describes an increasing sequence, while this sequence is decreasing. Always divide a later term by the earlier term to correctly find the ratio, including its sign.

The formula \(a_n = a_1 r^{n-1}\) correctly reflects that the ratio \(r\) is applied \(n-1\) times after the first term. The formula "\(a_n = a_1 + (n-1)d\)" is wrong here because it uses a common difference \(d\), which applies to arithmetic sequences, not geometric ones. Remember that geometric sequences multiply repeatedly by \(r\), while arithmetic sequences add repeatedly by \(d\).

The index of summation is the variable that changes value with each term, which here is \(i\). The number "\(5\)" is wrong because it is the upper limit, not the variable being incremented. In sigma notation, the letter beneath and above the \(\sum\) symbol identifies the index variable that counts each term.

Each term decreases by \(5\), so the next term after \(35\) is \(35 - 5 = 30\). The value "\(25\)" is wrong because it subtracts \(10\) instead of the actual common difference of \(5\). Always confirm the common difference using at least two pairs of consecutive terms before extending the pattern.

The common ratio is \(-2\), so multiplying the last term \(-24\) by \(-2\) gives \(48\). The value "\(-48\)" is wrong because it applies a ratio of \(2\) instead of \(-2\), ignoring the sign alternation in the sequence. When a geometric sequence alternates signs, the common ratio must be negative, and this sign must be carried through every multiplication.

The first term of a sequence is simply the value listed first, which is \(5\) here. The value "\(3\)" is incorrect because that is the common difference, not the starting term. Do not confuse the common difference \(d\) with the initial term \(a_1\); they play different roles in the sequence formula.

Each term is multiplied by \(\frac{1}{2}\) to get the next term, which is the defining property of a geometric sequence. It is not arithmetic because the differences between terms, such as \(5 - 10 = -5\) and \(2.5 - 5 = -2.5\), are not constant. Checking both the difference and the ratio between terms is the reliable way to classify any sequence.

Subtracting consecutive terms, \(-5 - (-2) = -3\), shows the common difference is \(-3\). The value "\(3\)" is wrong because it ignores that the sequence is decreasing, not increasing. Always subtract the earlier term from the later term in the correct order to preserve the sign of the difference.

Dividing \(-3\) by \(1\) gives \(-3\), and this ratio also correctly produces \(9\) and \(-27\) when applied repeatedly. The value "\(3\)" is wrong because it would produce all positive terms, not the alternating signs seen in this sequence. When a geometric sequence alternates in sign, the common ratio itself must be negative.

The lower limit of summation is the starting value written beneath the sigma symbol, which is \(3\) in this expression. The value "\(8\)" is incorrect because that number is the upper limit, marking where the summation ends. The lower and upper limits together define the full range of index values used to generate each term of the sum.

Using \(a_n = a_1 + (n-1)d\), \(a_{12} = 7 + (11)(-2) = 7 - 22 = -15\). The value "\(-17\)" is wrong because it incorrectly multiplies \(d\) by \(12\) instead of \(n-1 = 11\). Always use \(n-1\), not \(n\), as the multiplier for the common difference in the arithmetic term formula.

Using \(a_n = a_1 r^{n-1}\), \(a_6 = 5 \cdot 2^5 = 5 \cdot 32 = 160\). The value "\(80\)" is wrong because it uses \(2^4\) instead of the correct exponent \(2^5\). Remember that the exponent on \(r\) is always \(n-1\), one less than the term number being found.

Using \(S_n = \frac{n}{2}(2a_1 + (n-1)d)\), \(S_8 = \frac{8}{2}(4 + 35) = 4(39) = 156\). The value "\(150\)" is wrong because it results from an arithmetic slip when combining \(2a_1\) and \((n-1)d\) before multiplying. Always compute \(2a_1 + (n-1)d\) fully before multiplying by \(\frac{n}{2}\) to avoid order-of-operations errors.

Expanding the sum gives \(3 + 5 + 7 + 9 = 24\), found by substituting \(i = 1, 2, 3, 4\) into \(2i+1\). The value "\(20\)" is wrong because it likely omits one term or miscalculates \(2i+1\) for one of the index values. When evaluating sigma notation directly, it is safest to list every term before adding to avoid skipping a value of the index.

Adding the integers \(1 + 2 + 3 + 4 + 5\) gives \(15\), matching the known formula \(\frac{n(n+1)}{2}\) for \(n = 5\). The value "\(10\)" is wrong because it corresponds to the sum through \(k=4\), missing the final term. Using the formula \(\frac{n(n+1)}{2}\) is a fast check for sums of consecutive integers starting at \(1\).

Using \(a_{10} = a_1 + 9d\), \(42 = 6 + 9d\) gives \(9d = 36\), so \(d = 4\). The value "\(3\)" is wrong because it comes from dividing \(36\) by \(12\) instead of the correct multiplier \(9\). Remember that the multiplier on \(d\) is always \(n-1\), so for the 10th term it is \(9\), not \(10\).

Using \(a_4 = a_1 r^3\), \(32 = 4r^3\) gives \(r^3 = 8\), so \(r = 2\). The value "\(4\)" is wrong because it would give \(a_4 = 4 \cdot 4^3 = 256\), far larger than the given value. Always isolate \(r^{n-1}\) before taking the appropriate root to solve for the common ratio.

Using \(a_n = a_1 + (n-1)d\), \(51 = 3 + 4(n-1)\) gives \(48 = 4(n-1)\), so \(n - 1 = 12\) and \(n = 13\). The value "\(12\)" is wrong because it stops at \(n-1\) without adding \(1\) back to find the actual term number. After solving for \(n-1\), always remember to add \(1\) to get the final value of \(n\).

Using \(S_n = \frac{a_1(r^n - 1)}{r - 1}\), \(S_6 = \frac{3(64 - 1)}{1} = 3(63) = 189\). The value "\(192\)" is wrong because it likely uses \(r^6 = 64\) incorrectly as \(65\) before subtracting \(1\). Always compute \(r^n\) carefully first, then subtract \(1\), before multiplying by \(a_1\) and dividing by \(r-1\).

Using \(a_5 = a_1 + 4d\), \(20 = a_1 + 12\) gives \(a_1 = 8\). The value "\(5\)" is wrong because it results from subtracting \(d\) five times instead of the correct four times, since the multiplier is \(n-1\). When working backward from a later term, always use \(n-1\) steps of \(d\) to reach the first term.

Substituting \(n=1,2,3,4,5\) into \(4n-1\) gives \(3, 7, 11, 15, 19\), matching the series exactly. The expression "\(\sum_{n=0}^{4}(4n-1)\)" is wrong because at \(n=0\) it produces \(-1\), which is not the first term of the series. When writing sigma notation, always verify the formula by plugging in the starting index value first.

The formula \(a_n = 3n - 2\) is linear in \(n\), and consecutive terms always differ by the constant \(3\), which is the hallmark of an arithmetic sequence. It is not geometric because the ratio between terms, such as \(\frac{a_2}{a_1} = \frac{4}{1} = 4\) and \(\frac{a_3}{a_2} = \frac{7}{4}\), is not constant. A linear formula in \(n\) always signals an arithmetic sequence, while an exponential formula in \(n\) signals a geometric one.

The formula \(a_n = 2 \cdot 3^n\) has \(n\) in the exponent, meaning each term is \(3\) times the previous one, which is the definition of a geometric sequence. It is not arithmetic because the differences between terms, such as \(a_2 - a_1 = 18 - 6 = 12\) and \(a_3 - a_2 = 54 - 18 = 36\), are not constant. An exponential formula with \(n\) in the exponent always indicates a geometric sequence.

The arithmetic mean of two terms is their average, \(\frac{4+16}{2} = 10\), which keeps the difference constant at \(6\) on each side. The value "\(8\)" is wrong because it does not produce equal differences, since \(8-4=4\) but \(16-8=8\). Finding an arithmetic mean between two numbers always means taking their simple average.

The geometric mean of two positive numbers is \(\sqrt{4 \cdot 16} = \sqrt{64} = 8\), which keeps the ratio constant at \(2\) on each side. The value "\(10\)" is wrong because it is the arithmetic mean, not the geometric mean, and produces unequal ratios of \(2.5\) and \(1.6\). Remember that a geometric mean uses a square root of a product, unlike an arithmetic mean which uses a simple average.

Expanding the sum gives \(2^1 + 2^2 + 2^3 = 2 + 4 + 8 = 14\). The value "\(16\)" is wrong because it likely comes from adding an extra term, \(2^4 = 16\), beyond the stated upper limit of \(3\). Always double-check that the last term used matches the stated upper limit exactly, no more and no less.

This recursive rule adds \(5\) each time, making it arithmetic with \(d = 5\), so \(a_7 = 2 + 6(5) = 32\). The value "\(27\)" is wrong because it only adds \(5\) five times instead of the correct six times needed to reach the 7th term from the 1st. When converting a recursive rule to a direct formula, always use \(n-1\) additions of the common difference.

Applying the rule step by step, \(a_2 = 6\), \(a_3 = 12\), and \(a_4 = 24\), since each term doubles the previous one. The value "\(48\)" is wrong because it doubles one time too many, effectively computing \(a_5\) instead of \(a_4\). When working through a recursive definition, count carefully to make sure the number of multiplications matches the term you need.

This is an arithmetic sequence with \(a_1 = 20\) and \(d = 2\), so \(a_{15} = 20 + 14(2) = 20 + 28 = 48\). The value "\(46\)" is wrong because it uses \(13\) increases instead of the correct \(14\), since row 15 is \(14\) rows after row 1. In real-world arithmetic sequence problems, always subtract \(1\) from the term number to find the correct number of common-difference increases.

Using \(S_n = \frac{n}{2}(2a_1 + (n-1)d)\), \(S_{15} = \frac{15}{2}(-8 + 84) = 7.5(76) = 570\). The value "\(564\)" is wrong because it likely results from a slight arithmetic slip when combining \(2a_1\) and \((n-1)d\) before multiplying by \(\frac{n}{2}\). Always simplify the entire expression inside the parentheses completely before multiplying by \(\frac{n}{2}\) in the sum formula.

Using \(S = \frac{a_1}{1-r}\), \(S = \frac{9}{1-(-\frac{1}{3})} = \frac{9}{\frac{4}{3}} = 6.75\). The value "\(6\)" is wrong because it fails to correctly convert the fraction \(\frac{4}{3}\) in the denominator, leading to an underestimate. The infinite geometric sum formula only works when \(|r| < 1\), and here \(|-\frac{1}{3}| < 1\), so convergence is guaranteed.

Since \(|r| = 1.5\) is greater than \(1\), the terms grow without bound, so the series diverges by the convergence rule for infinite geometric series. The option "Converges" is wrong because convergence only occurs when \(|r| < 1\), which is not satisfied here. Always check that \(|r| < 1\) before attempting to apply the infinite geometric sum formula, since it is undefined otherwise.

Splitting the sum, \(3\sum_{i=1}^{6} i - 2(6) = 3(21) - 12 = 63 - 12 = 51\). The value "\(48\)" is wrong because it likely misapplies the constant term, subtracting \(2\) only once instead of \(6\) times, once for each of the six terms. When splitting a sum into separate pieces, remember that a constant term must be multiplied by the total number of terms in the sum.

The total distance equals the initial drop plus twice the infinite sum of the bounce heights, \(10 + 2\left(\frac{6}{1-0.6}\right) = 10 + 2(15) = 40\) feet. The value "\(25\) feet" is wrong because it only accounts for the bounce heights without doubling them to include both the upward and downward travel of each bounce. Bouncing-ball problems require doubling the infinite geometric sum of bounce heights since the ball travels up and back down each time, except for the very first drop.

Using \(S_n = \frac{a_1(r^n-1)}{r-1}\), \(242 = \frac{2(3^n-1)}{2}\) simplifies to \(3^n - 1 = 242\), so \(3^n = 243 = 3^5\), giving \(n = 5\). The value "\(4\)" is wrong because \(3^4 = 81\), which does not satisfy \(3^n = 243\). When solving for \(n\) in an exponential equation, rewrite both sides with the same base to directly compare exponents.

For an arithmetic sequence, the middle term equals the average of its neighbors, so \(2(x+1) = (x-1)+(2x+4)\) simplifies to \(2x+2 = 3x+3\), giving \(x = -1\). The value "\(1\)" is wrong because substituting it gives terms \(0, 2, 6\), whose differences \(2\) and \(4\) are not equal, so it fails the arithmetic condition. Setting the middle term equal to the average of the outer terms is the standard technique for solving arithmetic sequence variable problems.

For a geometric sequence, the middle term squared equals the product of its neighbors, so \(x^2 = 2 \cdot 18 = 36\), giving \(x = 6\) as the positive solution. The value "\(9\)" is wrong because substituting it gives ratios of \(4.5\) and \(2\), which are not equal, so the sequence would not be geometric. Setting the square of the middle term equal to the product of the outer terms is the standard technique for solving geometric sequence variable problems.

This is a geometric sequence with \(a_1 = 40{,}000\) and \(r = 1.10\), so \(a_4 = 40{,}000(1.10)^3 = 40{,}000(1.331) = 53{,}240\). The value "\(52{,}000\)" is wrong because it results from a linear estimate of adding a flat \(4{,}000\) each year rather than compounding the 10% growth. Percent-growth problems like salary raises are modeled as geometric sequences, not arithmetic ones, because the growth compounds multiplicatively each period.

Splitting the sum, \(5\sum_{k=1}^{10} k - 3(10) = 5(55) - 30 = 275 - 30 = 245\). The value "\(240\)" is wrong because it likely subtracts \(35\) instead of the correct constant total of \(30\), miscounting how many times \(3\) should be subtracted. When a sum includes a constant term, that constant must be multiplied by the number of terms, which here is \(10\), not the value of \(k\).

Using \(a_n = a_1 r^{n-1}\), \(512 = 4(2)^{n-1}\) gives \(2^{n-1} = 128 = 2^7\), so \(n - 1 = 7\) and \(n = 8\). The value "\(7\)" is wrong because it stops at solving for \(n-1\) without adding \(1\) back to find the actual term count. After solving the exponent equation for \(n-1\), always remember to add \(1\) to find the true number of terms.

Using \(S_n = \frac{n}{2}(a_1 + a_n)\), \(1683 = \frac{n}{2}(102)\) gives \(1683 = 51n\), so \(n = 33\). The value "\(30\)" is wrong because \(51(30) = 1530\), which does not match the given sum of \(1683\). When both the first and last terms are known, the sum formula \(S_n = \frac{n}{2}(a_1+a_n)\) is often the fastest way to solve for \(n\) without finding \(d\) first.

This is an infinite geometric series with \(a_1 = 5\) and \(r = 0.8\), and since \(|0.8| < 1\) it converges to \(S = \frac{5}{1-0.8} = \frac{5}{0.2} = 25\). The value "\(20\)" is wrong because it likely comes from dividing \(5\) by \(0.25\) instead of the correct denominator \(0.2\), misapplying the formula \(1-r\). Always verify \(|r|<1\) first, then carefully compute \(1-r\) before dividing \(a_1\) by it to find the infinite sum.