Recognize a sequence’s structure, then choose pairing, telescoping, a recurrence or a justified limit to sum it.
JEE Main · JEE Advanced80 min study guide11 indexed sections
BY THE END, YOU CAN
Derive and apply arithmetic and geometric term/sum formulas with exceptional ratios handled.
Separate a sequence of terms from its sequence of partial sums.
Evaluate telescoping and arithmetic-geometric sums.
Solve elementary recurrences and test proposed term formulas.
Distinguish finite sums from convergent infinite series.
BEFORE YOU START
Algebra, powers and finite summation notation.
Basic limits and logarithms.
Factorization and partial fractions.
Original educational lesson. Work through the examples before revealing the practice answers. Check the current official syllabus separately; historical question frequency does not establish eligibility or predict future questions.
CONNECT THE CONCEPT TO THE EXAM
Selected exam questions & worked solutions
These selected examples are separate from the original practice below. Each identifies its exam, year and wording. They do not represent every paper or every selected year.
Available in this lesson: JEE Advanced 2014. This is a limited lesson sample, not a complete exam-question bank.
1 sourced sample question available across the listed exams and years.
Matched topics: Arithmetic and Geometric Progressions.
Independently derived and checked solution. Archived key finality is unverified; this is not presented as an official final answer.
Positive integers a, b, c form a geometric progression with integer ratio b/a. Their arithmetic mean is b + 2. Evaluate (a² + a − 14)/(a + 1).
Reveal answer & worked solution
Answer: 4; independently derived and consistent with the archived answer annotation.
Put b = ar and c = ar², where r is a positive integer. The mean condition gives a(1+r+r²) = 3ar + 6.
Rearrange: a(r−1)² = 6. The case r=1 is impossible; since (r−1)² is a positive square dividing 6, it equals 1.
With positive integer r, this gives r=2 and a=6. Substitute: (36+6−14)/7 = 4.
Source: JEE (Advanced) 2014, Paper 1, Code 1, Q51; solution independently derived; archived-key finality unverified. Solution written for this lesson.
01 / UNDERSTAND & APPLY
1. A sequence is an ordered rule; a series is a sum
A sequence assigns a value aₙ to each positive integer n, or to another explicitly stated index set. The index tells you the position, not the value. A series adds terms of a sequence, and its nth partial sum is Sₙ = a₁ + ⋯ + aₙ. The terms and the partial sums are different sequences and can behave very differently.
An explicit formula gives aₙ directly from n. A recurrence gives later terms using earlier ones and needs enough initial values to determine a particular sequence. A few displayed terms can suggest a pattern but cannot logically determine a unique infinite sequence without an additional rule. Always distinguish an inferred pattern from a specified definition.
An arithmetic progression has constant successive difference d: aₙ₊₁ − aₙ = d. A geometric progression has a constant multiplier r: aₙ₊₁ = r aₙ. When terms are nonzero, the ratio aₙ₊₁/aₙ equals r, but the multiplier definition also handles zero terms without division. Neither progression must consist of positive or increasing terms.
An AP can decrease if d < 0 or be constant if d = 0. A GP can alternate signs if r < 0, shrink in magnitude if |r| < 1, or vanish after its first term if r = 0. Knowing the type of progression is more useful than assuming a familiar-looking monotone picture.
02 / UNDERSTAND & APPLY
2. Arithmetic formulas come from repeated addition and pairing
For an AP with first term a and difference d, repeated addition gives aₙ = a + (n−1)d. The offset is n−1 because the first term already occurs before any step. If two terms aₚ and a_q are known with p ≠ q, their difference is (q−p)d, allowing d to be determined before solving for a.
To derive the sum, write Sₙ forward and backward. Each paired position sums to a + aₙ, and there are n pairs. Thus 2Sₙ = n(a + aₙ), giving Sₙ = n[2a + (n−1)d]/2. This argument works whether d is positive, negative or zero; no positivity is required.
If a term number must be found from a target value, solve the linear equation for n and check that n is a positive integer. A fractional or negative index is not an allowed term of a sequence indexed by positive integers. If a sum condition produces a quadratic in n, test every root against the same index restriction and the original equation.
Three numbers a,b,c form an AP exactly when 2b = a + c. The middle term is the arithmetic mean of the endpoints. This relation remains valid for negative or zero values. It does not imply the numbers are written in increasing order.
ORIGINAL WORKED EXAMPLE
Recover an AP from two positions
An AP has fourth term 11 and ninth term 26. Find its first term, common difference and sum of the first twelve terms.
The equations are a + 3d = 11 and a + 8d = 26. Subtracting gives 5d = 15, so d = 3.
Then a = 11 − 9 = 2. The twelfth term is 2 + 11·3 = 35.
The sum is S₁₂ = 12(2 + 35)/2 = 6·37 = 222.
Check the original positions: 2 + 3·3 = 11 and 2 + 8·3 = 26.
Answer a = 2, d = 3, S₁₂ = 222.
03 / UNDERSTAND & APPLY
3. Subtract a shifted GP to cancel its middle terms
For first term a and ratio r, the nth term is ar^(n−1). The finite sum is Sₙ = a + ar + ⋯ + ar^(n−1). Multiplying by r shifts all powers up one. Subtract rSₙ from Sₙ to obtain (1−r)Sₙ = a(1−rⁿ). Hence Sₙ = a(1−rⁿ)/(1−r) for r ≠ 1.
The case r = 1 is not undefined as a sequence; every term equals a and Sₙ = na. It is only the divided formula that is unavailable. For r = 0 and n ≥ 1, the terms are a,0,0,… and the sum is a. Writing out these special cases prevents accidental use of 0⁰ conventions at the first term.
If all terms are nonzero, three consecutive GP terms satisfy b² = ac. Conversely, when a ≠ 0 and b² = ac, setting r = b/a gives the required third term c = ar². If a = 0, this shortcut needs separate checking: the multiplier definition may force more zeros than the single equation b² = ac reveals.
When a ratio is negative, signs alternate. The finite sum formula remains valid, but absolute values or geometric intuition based on positive bars may fail. A quick list of the first few terms is a good sign check before evaluating a large power.
ORIGINAL WORKED EXAMPLE
A finite GP with alternating signs
Find the sum of the first six terms of a GP with first term 3 and ratio −2.
The six terms are 3,−6,12,−24,48,−96.
Use r ≠ 1: S₆ = 3[1−(−2)⁶]/[1−(−2)] = 3(1−64)/3.
The value is −63.
Direct addition in pairs gives (3−6)+(12−24)+(48−96)=−3−12−48=−63, checking the alternating signs.
Answer −63.
04 / UNDERSTAND & APPLY
4. An infinite sum is a limit of finite sums
The infinite series a + ar + ar² + ⋯ means the limit of its partial sums, not an arithmetic operation performed on an infinitely long completed list. For |r| < 1, rⁿ tends to zero, so the finite GP formula tends to a/(1−r). This is the convergence condition behind the familiar expression.
If a ≠ 0 and |r| ≥ 1, the terms arⁿ do not tend to zero, so the series cannot converge. At r = −1, partial sums oscillate; at r = 1, nonzero constant terms accumulate; for larger magnitude ratios the terms grow. The all-zero series a = 0 is a trivial exception and converges for any specified multiplier.
The remainder after n terms of a convergent GP is Rₙ = arⁿ/(1−r). Its absolute value gives the error in replacing the infinite sum by the finite partial sum. With a negative ratio, the remainder’s sign alternates, but the absolute error bound remains straightforward.
A convergent sequence of terms need not produce a convergent series. The terms 1/n tend to zero, but the harmonic series diverges. Tending to zero is necessary, not sufficient, for an infinite series. Avoid using the GP formula on a sequence merely because its terms decrease.
ORIGINAL WORKED EXAMPLE
A sum and a controlled truncation error
Find the sum of 6 + 2 + 2/3 + ⋯, and the smallest n for which the absolute remainder after n terms is below 1/100.
This is a GP with a=6 and r=1/3, so its sum is 6/(1−1/3)=9.
The remainder is Rₙ=6(1/3)ⁿ/(2/3)=9/3ⁿ.
The condition 9/3ⁿ < 1/100 is 3ⁿ > 900. Since 3⁶=729 and 3⁷=2187, the smallest n is 7.
The strict inequality matters: an error exactly equal to the threshold would not satisfy “below.” Here the two neighboring powers settle the integer choice exactly.
Answer Infinite sum 9; seven terms are needed.
05 / UNDERSTAND & APPLY
5. Recover terms from partial sums and shift indices carefully
If Sₙ is known, then aₙ = Sₙ − Sₙ₋₁ for n ≥ 2, and a₁ = S₁. One may extend the difference formula to n=1 only if S₀ is consistently defined as zero. A polynomial expression for Sₙ given only for positive n should not be assumed valid at zero without checking.
Finite sums are linear: Σ(αaₖ + βbₖ) = αΣaₖ + βΣbₖ over the same index range. A change of index must change both bounds. For example, Σₖ₌₁ⁿ(k+1)² = Σⱼ₌₂^(n+1)j². The missing first term and added last term are not negligible in an exact finite calculation.
The standard power sums are Σk = n(n+1)/2, Σk² = n(n+1)(2n+1)/6, and Σk³ = [n(n+1)/2]². They can be derived by telescoping differences of successive powers or by induction. Their degree in n is one higher than the summed power, which provides a useful leading-order check.
For instance, (k+1)³−k³ = 3k²+3k+1. Summing both sides makes the left side telescope, and using the known sum of k then solves for Σk². This shows why summation formulas are connected to finite differences, the discrete counterpart of differentiation.
ORIGINAL WORKED EXAMPLE
A partial-sum rule determines an AP
A sequence has Sₙ = 2n² + 3n for positive integers n. Find aₙ and identify its type.
The first term is a₁=S₁=5. The same polynomial has value zero at n=0, so extending S₀=0 is consistent.
For n≥1, aₙ=Sₙ−Sₙ₋₁ = (2n²+3n)−[2(n−1)²+3(n−1)].
Expand and simplify to aₙ=4n+1.
The terms are 5,9,13,…, an AP with common difference 4. Its standard sum formula gives n[2·5+(n−1)4]/2=2n²+3n, checking the reconstruction.
Answer aₙ = 4n + 1; an AP with first term 5 and difference 4.
06 / UNDERSTAND & APPLY
6. Expose cancellation before evaluating terms
A telescoping sum has terms expressible as bₖ − bₖ₊₁, so most intermediate contributions cancel. The finite result is b₁ − bₙ₊₁. Write the first few and last few terms explicitly to see the surviving boundaries. Assuming everything cancels to zero usually loses exactly those boundary terms.
Partial fractions often create a telescoping form. The identity 1/[k(k+1)] = 1/k − 1/(k+1) leaves 1−1/(n+1) after summing from 1 to n. A shifted gap, such as 1/[k(k+2)], leaves more than one boundary term, so do not reuse the one-step result unchanged.
Telescoping also works with radicals after rationalization. Since √(k+1)−√k = 1/[√(k+1)+√k], a sum of reciprocal adjacent-root sums collapses to a difference of endpoint roots. The algebraic identity should be checked before summing; a missing sign or factor changes every term.
For an infinite telescoping series, first obtain the exact finite partial sum and then take its limit. This avoids informal cancellation of infinite expressions and makes convergence or divergence visible through the surviving boundary term.
ORIGINAL WORKED EXAMPLE
A two-step telescoping pattern
Evaluate Σₖ₌₁ⁿ1/[k(k+2)] for positive integer n.
Partial fractions give 1/[k(k+2)] = (1/2)(1/k − 1/(k+2)).
The sum is half of [1 + 1/2 + ⋯ + 1/n − (1/3 + ⋯ + 1/(n+2))].
All common middle terms cancel, leaving (1/2)[1 + 1/2 − 1/(n+1) − 1/(n+2)].
Thus the result is 3/4 − 1/[2(n+1)] − 1/[2(n+2)]. At n=1 it gives 1/3, matching the single original term.
Answer 3/4 − 1/[2(n+1)] − 1/[2(n+2)].
07 / UNDERSTAND & APPLY
7. Shift a weighted GP to reduce the weight
An arithmetic-geometric sum has a linearly changing coefficient multiplying geometric powers, such as S = 1 + 2r + 3r² + ⋯ + nr^(n−1). Multiplying by r and subtracting reduces the coefficients: (1−r)S = 1 + r + ⋯ + r^(n−1) − nrⁿ. The remaining ordinary GP can then be summed.
For r ≠ 1, this gives S = [1−(n+1)rⁿ+nr^(n+1)]/(1−r)². At r=1, use S=n(n+1)/2 directly. The finite formula is valid for negative ratios too; convergence restrictions arise only when taking n to infinity.
For |r|<1, the terms nrⁿ tend to zero and the infinite weighted sum is 1/(1−r)². A derivative of the finite GP polynomial gives the same finite identity and serves as an independent derivation. Differentiating an infinite series requires justification, so deriving finite sums first is a safe elementary route.
More complicated polynomial weights can be reduced by repeated finite differences or repeated differentiation of a generating polynomial. Before expanding a large expression, ask whether shifting the index or multiplying by the ratio will simplify the coefficient pattern.
ORIGINAL WORKED EXAMPLE
A weighted geometric sum
Evaluate Σₖ₌₁ⁿ k/2ᵏ, and then its infinite limit.
Let Sₙ=1/2+2/2²+⋯+n/2ⁿ. This is one-half of the standard weighted sum with r=1/2.
Substituting into the finite formula gives Sₙ = 2 − (n+2)/2ⁿ.
Check n=1: 2−3/2=1/2, the single term. For n=2, the formula gives 2−4/4=1, agreeing with 1/2+2/4.
As n tends to infinity, (n+2)/2ⁿ tends to zero, so the infinite series sums to 2.
Answer Sₙ = 2 − (n+2)/2ⁿ; infinite sum 2.
08 / UNDERSTAND & APPLY
8. A recurrence can hide a GP after a shift
For a recurrence aₙ₊₁ = raₙ + c with r ≠ 1, a fixed point L satisfies L=rL+c, hence L=c/(1−r). Subtract it: aₙ₊₁−L = r(aₙ−L). The shifted sequence is geometric, giving aₙ = L + (a₁−L)r^(n−1). This method explains the formula rather than treating it as another pattern to memorize.
If r=1, the recurrence is aₙ₊₁=aₙ+c, an AP. If |r|<1, the shifted term tends to zero and the sequence tends to L. If r<0, it may alternate around L. If |r|>1 and a₁≠L, deviations grow in magnitude. The exceptional initial value a₁=L gives a constant sequence for any r.
To verify an explicit recurrence solution, check the initial value and substitute into the recurrence. Checking only the first few terms is not a proof for all n. Induction packages these two checks: the base case establishes the starting term, and the recurrence establishes the step from n to n+1.
A monotone bounded sequence has a finite limit, but substituting a guessed limit into a recurrence only identifies possible limits; it does not prove convergence. First establish convergence by a valid argument, or use an explicit formula whose limiting behavior is known.
ORIGINAL WORKED EXAMPLE
Shift to the fixed point
A sequence satisfies a₁=1 and aₙ₊₁=(1/2)aₙ+3. Find aₙ and its limit.
Therefore aₙ=6−5/2^(n−1). At n=1 this gives 1, as required.
Substitution into the recurrence reproduces the formula with n replaced by n+1. Since 2^(−(n−1)) tends to zero, the limit is 6.
Answer aₙ = 6 − 5/2^(n−1); limit 6.
09 / UNDERSTAND & APPLY
9. Original practice: APs and GPs
Keep the index domain explicit and treat exceptional ratios separately. A sum and a term with the same subscript are not interchangeable.
ORIGINAL PRACTICE · 1
Find the twentieth term of an AP with a₁=7 and d=−2.
Reveal answer & reasoning
−31.
Use a₂₀=7+19(−2)=−31. The offset is 19 steps, not 20.
ORIGINAL PRACTICE · 2
Find the sum of the first ten positive odd integers.
Reveal answer & reasoning
100.
The AP has first term 1 and tenth term 19. Its sum is 10(1+19)/2=100.
ORIGINAL PRACTICE · 3
An AP has a₃=8 and a₇=20. Find a₁.
Reveal answer & reasoning
2.
Subtracting gives 4d=12, so d=3. Then a₁=a₃−2d=8−6=2.
ORIGINAL PRACTICE · 4
Find the fifth term of the GP 2,6,18,….
Reveal answer & reasoning
162.
The ratio is 3, so a₅=2·3⁴=162.
ORIGINAL PRACTICE · 5
Find the sum of eight terms of the constant GP 4,4,4,….
Reveal answer & reasoning
32.
Here r=1, so use S₈=8·4. The divided formula with denominator 1−r is not applicable.
ORIGINAL PRACTICE · 6
Does 1−1+1−1+⋯ converge in the ordinary partial-sum sense?
Reveal answer & reasoning
No.
The partial sums alternate between 1 and 0, so they have no single limit. Substituting r=−1 into the convergent-GP formula would violate |r|<1.
ORIGINAL PRACTICE · 7
Find the infinite sum 5−5/2+5/4−⋯.
Reveal answer & reasoning
10/3.
The ratio is −1/2, whose magnitude is below one. Thus the sum is 5/[1−(−1/2)]=5/(3/2)=10/3.
10 / UNDERSTAND & APPLY
10. Original practice: differences, cancellation and recurrence
For a finite sum, retain all endpoint terms. For an infinite sum, take the limit of a proven finite formula rather than cancelling an infinite list informally.
ORIGINAL PRACTICE · 1
If Sₙ=n² for n≥1, find aₙ.
Reveal answer & reasoning
aₙ=2n−1.
With S₀=0 consistent, aₙ=n²−(n−1)²=2n−1. These are the positive odd integers.
ORIGINAL PRACTICE · 2
Evaluate Σₖ₌₁ⁿ1/[k(k+1)].
Reveal answer & reasoning
n/(n+1).
Write the term as 1/k−1/(k+1). The finite sum telescopes to 1−1/(n+1)=n/(n+1).
ORIGINAL PRACTICE · 3
Evaluate Σₖ₌₁ⁿ1/[√(k+1)+√k].
Reveal answer & reasoning
√(n+1)−1.
Rationalization makes each term √(k+1)−√k. The intermediate roots cancel, leaving the final root minus √1.
ORIGINAL PRACTICE · 4
Evaluate Σₖ₌₁⁵ k².
Reveal answer & reasoning
55.
Use n(n+1)(2n+1)/6 at n=5: 5·6·11/6=55. Direct addition 1+4+9+16+25 checks it.
ORIGINAL PRACTICE · 5
Solve a₁=2, aₙ₊₁=3aₙ−4.
Reveal answer & reasoning
aₙ=2 for every n.
The fixed point satisfies L=3L−4, so L=2. The initial value is exactly that fixed point, leaving zero geometric deviation despite the ratio magnitude exceeding one.
ORIGINAL PRACTICE · 6
Solve a₁=3, aₙ₊₁=2aₙ+1.
Reveal answer & reasoning
aₙ=2^(n+1)−1.
The fixed point is −1. Thus aₙ+1=(a₁+1)2^(n−1)=4·2^(n−1)=2^(n+1). Subtract one and check the initial value.
ORIGINAL PRACTICE · 7
If aₙ→0, must Σaₙ converge?
Reveal answer & reasoning
No.
The harmonic terms aₙ=1/n tend to zero, but their partial sums are unbounded. A zero term limit is necessary for series convergence, not sufficient.
11 / UNDERSTAND & APPLY
11. Choose the operation that exposes structure
For an AP, subtract successive terms or pair endpoints. For a GP, multiply the sum by its ratio and subtract. For a telescoping sum, express each term as a difference. For an arithmetic-geometric sum, shift to reduce the coefficient weight. For an affine recurrence, subtract the fixed point. These are variations on the same idea: transform the expression so that most of it cancels.
The common errors are confusing aₙ with Sₙ, using n instead of n−1 in a term formula, ignoring the r=1 case, applying an infinite GP sum outside |r|<1, dropping boundary terms in telescoping, and accepting a noninteger term index. Another common gap is finding a candidate limit from a recurrence without proving the sequence converges.
Check an exact formula at n=1 and one further small n, then verify its recurrence or finite-difference relation. For parameter conditions, test exceptional values separately. For an infinite result, state the convergence condition and show the remaining term tends to zero.
A few numerical terms are useful for detecting a sign or indexing error, but they are not the proof. The proof is the algebraic identity, the induction step, or the justified limiting argument that applies to every allowed index.
KEEP THE BIG IDEAS
Quick revision
APs have a constant difference; GPs have a constant multiplier, including zero and negative cases.
Finite GP formulas and infinite GP limits have different conditions.
Terms are differences of consecutive partial sums.
Telescoping leaves boundary terms; weighted GPs simplify by shifting.
A recurrence’s fixed point suggests a transformation, but convergence still needs justification.
The teaching notes, worked explanations and original practice are written for this lesson. Separately labeled selected exam questions identify their exam, year, source and whether wording is paraphrased or an exact excerpt. Educational references do not certify current exam coverage.