Geometric sequence homework: keep the ratio

Geometric sequence homework starts with a common ratio

A geometric sequence is formed by repeatedly multiplying the preceding term by one fixed ratio q. West Texas A&M’s lesson on geometric sequences distinguishes that multiplier from an arbitrary changing pattern. For 6, −12, 24, −48, the successive quotients are −12/6 = −2, 24/(−12) = −2 and −48/24 = −2. Thus the displayed terms are consistent with q = −2. The corresponding differences are not constant, so an arithmetic-sequence rule would not fit.

For ratio calculations in our worked examples, the first term and ratio are nonzero, so all the denominators are nonzero. Never use a quotient with a zero denominator. A recurrence can describe special cases involving zeros, but those need separate reasoning. Also, a few displayed terms do not prove a unique infinite continuation unless the problem specifies a geometric model. For comparison, 6, 12, 18 has constant differences of 6 but ratios 2 and 1.5, so it is not geometric.

Count multiplications in the term formula

If the first term is labelled a_1, then a_2 = a_1q, a_3 = a_1q² and a_4 = a_1q³. Reaching term n from term 1 takes n − 1 multiplications, so a_n = a_1q^(n − 1), with n a positive integer. For our example, a_n = 6(−2)^(n − 1). Substituting n = 1 must return 6 because a nonzero base to power zero is 1. That is a quick indexing check.

For n = 4, calculate 6(−2)³ = 6(−8) = −48. Using exponent 4 would produce 96, which is the fifth term. If a problem instead starts with a_0 = 6, its formula is a_n = 6(−2)^n for n ≥ 0. Read the starting index before using a memorized template. The same numerical progression can have different labels; changing a label must not silently move every value one position along the list.

Keep the entire negative ratio inside parentheses

The expression (−2)^k includes the negative sign in the base. Even powers are positive and odd powers negative. In 6(−2)^(n − 1), the sign therefore alternates as n changes. By contrast, −2^k conventionally means −(2^k), with the minus outside the power. For k = 2, (−2)² = 4 but −2² = −4. Losing parentheses changes the rule, not just its appearance on the page or calculator.

Trace the next two terms by recurrence as a check: −48 multiplied by −2 gives 96; 96 multiplied by −2 gives −192. The term formula gives the same fifth and sixth terms. Our guide to exponent rules explains why the base and exponent boundaries matter. An alternating sequence need not decrease: here the magnitudes double while the signs switch. Describe sign and magnitude separately rather than calling the whole list increasing simply because its absolute values get larger.

Nonconsecutive known terms can leave two possible ratios

For a geometric sequence with nonzero terms, a_j/a_i = q^(j − i). If a_3 = 24 and a_6 = −192, there are three multiplication steps, so q³ = −192/24 = −8. Over the real numbers this gives q = −2. Recover the first term: a_1 = a_3/q² = 24/4 = 6. Check both original positions rather than accepting a ratio that fits only one of them.

Now consider a separate task giving only a_2 = −12 and a_4 = −48, with no first term specified. It gives q² = 4, so q can be +2 or −2. With q = +2, a_1 = −6 and a_3 = −24; with q = −2, a_1 = 6 and a_3 = 24. Both fit the two stated terms. An even index gap can hide the sign. Do not silently select the positive root unless an additional condition, such as a positive ratio, requires it.

A finite sum is different from the final term

The fourth term is a_4 = −48, but the sum of the first four terms is S_4 = 6 − 12 + 24 − 48 = −30. To derive a general finite sum, write S_n = a_1 + a_1q + … + a_1q^(n − 1). Multiplying this whole equation by q shifts the powers. Subtract the shifted equation from the first: (1 − q)S_n = a_1(1 − q^n). The middle terms cancel because they are the same terms with opposite signs.

For q ≠ 1, divide to obtain S_n = a_1(1 − q^n)/(1 − q). Here S_4 = 6[1 − (−2)^4]/[1 − (−2)] = 6(1 − 16)/3 = −30. Keep parentheses around the entire numerator and denominator. For q = 1, division by 1 − q is forbidden; every term is a_1 and S_n = na_1. A finite sum formula does not require |q| < 1, which is a condition for a different question about an infinite limit.

Do not use an infinite-sum formula without convergence

For 12, 6, 3, … the ratio is 1/2. The first three terms sum to 21, and S_n = 12[1 − (1/2)^n]/(1 − 1/2) = 24[1 − (1/2)^n]. As n grows, (1/2)^n tends to zero, so the partial sums approach 24. This is an infinite geometric sum defined by a limit of finite sums; it is not the value of a particular term. No finite partial sum in this positive example reaches 24 exactly.

For a nonzero first term, the infinite geometric sum exists when |q| < 1 and equals a_1/(1 − q). The absolute value is essential: q = −2 is less than 1, but its magnitude is 2. Our original sequence has growing magnitudes and no convergent infinite sum. Substituting 6/(1 − (−2)) = 2 would therefore be an invalid use of the limit formula. Its partial sums −6, 18 and −30 already warn that they are not settling near 2.

Ask Eqora about the first ratio or sign you cannot justify

Start with your own list of terms, the proposed ratio and the first doubtful formula. If your course permits AI assistance, capture one complete readable problem and the relevant attempt. Preserve term subscripts, a possible starting index of zero, parentheses and minus signs. Avoid personal information. Before inspecting proposed steps in Eqora, verify that a_3 = 24 and a_6 = −192 have not become adjacent terms or lost their indices. An explanation of a different captured problem cannot validate your original work.

Ask a focused follow-up such as: “Why does a_6/a_3 give q³ rather than q², and what determines the sign of q?” For the separate even-gap exercise, ask why both roots survive and what extra information would choose one. These are suggested learning prompts, not quotations from an app test. Our guide to asking focused follow-up questions explains how to name the uncertainty instead of requesting another complete result without identifying the confusing transition.

Check any suggested answer against the recurrence and the original known terms. Stop if it discards the negative ratio without justification, changes S_n into a_n or applies the infinite-sum formula to q = −2. Consult the algebra or your teacher’s method. These examples can be solved on paper without a paid service; that does not promise every app feature or unlimited use is free. Explain the corrected step in your own words, then close the assisted solution for the independent practice.

Change the numbers and check the recurrence independently

Try a new sequence with b_1 = 5 and q = 3. The first four terms are 5, 15, 45, 135. Thus b_4 = 135, whereas S_4 = 200. The formulas agree: 5 × 3³ = 135 and 5(1 − 3^4)/(1 − 3) = 200. Write both answers with their requested quantity. A correct number attached to the wrong symbol would still answer the wrong question, even if the arithmetic were flawless.

For a fractional ratio, use c_1 = 18 and q = 2/3. The first four terms are 18, 12, 8 and 16/3, so c_4 = 16/3 and their sum is 130/3. Keep the ratio as a fraction rather than rounding it repeatedly. Finally, decide which examples admit a convergent infinite sum: the ratio 3 does not, while 2/3 does and gives 18/(1 − 2/3) = 54. Confirm the multiplier, index, signs and requested quantity before relying on the final expression.