Quick Summary
Concept Tested: Summation of a series whose general term combines a constant with a geometric progression.
Chapter – Subtopic: Sequences and Series – Sum of Mixed (Arithmetic + Geometric) Series
Difficulty: ★★★☆☆ (3/5)
Estimated Time: 5 minutes
Key Formula: $$S_n=\sum_{k=1}^n\left(1+\frac{1}{3^k}\right)=n+\frac12\left(1-3^{-n}\right)$$
Answer: C
One-line Reason: Each term equals $1+3^{-k}$; splitting the sum into $n$ ones plus a finite geometric series gives $n+\frac12-\frac{1}{2\cdot3^n}$.
The Question
The sum of the series $$\frac43+\frac{10}{9}+\frac{28}{27}+\cdots\text{ upto }n\text{ terms}$$ is:
(A) $n-\dfrac12+\dfrac{1}{2\cdot3^n}$
(B) $n+\dfrac12+\dfrac{1}{2\cdot3^n}$
(C) $n+\dfrac12-\dfrac{1}{2\cdot3^n}$
(D) $n-\dfrac12-\dfrac{1}{2\cdot3^n}$
Quick Answer
Answer: C. Each term of the series has the form $1+\dfrac{1}{3^k}$. Summing the constant part gives $n$, and summing the geometric part ($\dfrac13+\dfrac19+\dfrac1{27}+\cdots$, $n$ terms) gives $\dfrac12(1-3^{-n})$. Adding these: $n+\dfrac12-\dfrac{1}{2\cdot3^n}$.
Why Other Options Are Incorrect
Option A: $n-\dfrac12+\dfrac{1}{2\cdot3^n}$ flips the sign of the constant $\dfrac12$ term — this would arise from mistakenly writing the geometric sum as $-\dfrac12(1-3^{-n})$ instead of $+\dfrac12(1-3^{-n})$.
Option B: $n+\dfrac12+\dfrac{1}{2\cdot3^n}$ has the correct constant term but the wrong sign on $3^{-n}$ — this results from expanding $(1-3^{-n})$ as $(1+3^{-n})$ by mistake.
Option D: $n-\dfrac12-\dfrac{1}{2\cdot3^n}$ has both signs wrong relative to C — consistent with a full sign inversion of the geometric-sum formula $\dfrac{a(1-r^n)}{1-r}$.
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Understanding the Concept
This series is not purely arithmetic or purely geometric — each term is a sum of a constant (1) and a geometric term ($3^{-k}$). The standard technique for such “mixed” series is to split the summation into two separate, simpler sums: a constant sum (trivial: $n$ copies of 1) and a standard finite geometric sum, then combine the results.
Key principle: $$\sum_{k=1}^n\left(1+\frac{1}{3^k}\right)=\underbrace{\sum_{k=1}^n1}_{n}+\underbrace{\sum_{k=1}^n\left(\frac13\right)^k}_{\text{finite GP}}$$
Detailed Step-by-Step Solution
Step 1: Identify the general term
Writing each term as a fraction over increasing powers of 3: $\dfrac43=1+\dfrac13$, $\dfrac{10}9=1+\dfrac19$, $\dfrac{28}{27}=1+\dfrac1{27}$. This confirms the general term is $$T_k=1+\frac{1}{3^k},\quad k=1,2,\dots,n$$
Step 2: Split the summation
$$S_n=\sum_{k=1}^nT_k=\sum_{k=1}^n1+\sum_{k=1}^n\left(\frac13\right)^k=n+\sum_{k=1}^n\left(\frac13\right)^k$$
Step 3: Sum the finite geometric series
The second sum is a GP with first term $a=\dfrac13$, common ratio $r=\dfrac13$, and $n$ terms:
$$\sum_{k=1}^n\left(\frac13\right)^k=\frac{a(1-r^n)}{1-r}=\frac{\frac13\left(1-3^{-n}\right)}{1-\frac13}=\frac{\frac13\left(1-3^{-n}\right)}{\frac23}=\frac12\left(1-3^{-n}\right)$$
Step 4: Combine
$$S_n=n+\frac12\left(1-3^{-n}\right)=n+\frac12-\frac{1}{2\cdot3^n}\ \rightarrow\ \textbf{Option C}$$
Final Answer
✔ The correct answer is (C) — $S_n=n+\dfrac12-\dfrac{1}{2\cdot3^n}$.
Essential Formulas for This Topic
$$\text{Finite GP sum: }\sum_{k=1}^n ar^{k-1}=\frac{a(1-r^n)}{1-r}\quad(r\neq1)$$
$$\text{Mixed-series split: }\sum(c+f(k))=nc+\sum f(k)\text{, for constant }c$$
Note: always separate the constant and variable parts of a term before summing — never try to fit a mixed term into a single GP formula directly.
Common Mistakes to Avoid
Mistake 1: Treating the whole series as a single geometric progression
Since $\frac43,\frac{10}9,\frac{28}{27}$ do not have a constant ratio ($\frac{10/9}{4/3}=\frac56\neq\frac{28/27}{10/9}=\frac{14}{15}$), applying the GP-sum formula directly to the original terms (without splitting off the “+1”) gives a wrong, non-existent common ratio.
Mistake 2: Sign error in the GP sum formula
Writing $\frac{a(1-r^n)}{1-r}$ as $\frac{a(1+r^n)}{1-r}$ or flipping $1-r$ to $r-1$ produces Options A, B, or D instead of C — always double-check the sign before simplifying.
Mistake 3: Forgetting to convert $n$ copies of 1 into just “$n$”
Some students keep re-deriving $\sum_{k=1}^n 1$ as a series instead of recognizing it’s simply $n$ — this leads to an unsimplified, harder-to-match final expression.
Key Concept Summary
- A series whose term is (constant + geometric part) should be split into two separate sums.
- The constant part sums to $n\times(\text{constant})$.
- The variable part is summed using the standard finite GP formula.
- Always verify the closed form by direct substitution for small $n$ (e.g., $n=1,2,3$).
- Watch the sign inside $(1-r^n)$ carefully — it is the single most common source of error here.
Golden Rule: When a series term is a sum of two different types of sequences, split the summation before applying any single formula.
Frequently Asked Questions
Q: Why can’t we just apply the GP-sum formula directly to $\frac43,\frac{10}9,\frac{28}{27},\dots$?
A: Because these terms do not share a common ratio — they are not a geometric progression at all. Only the “$3^{-k}$” part of each term is geometric; the “+1” part is constant.
Q: How do we know the general term is $1+3^{-k}$ and not something else?
A: By expressing each given term as a fraction with denominator $3^k$: $\frac43=\frac{3+1}{3}=1+\frac13$, and similarly for the rest — the pattern numerator $=3^k+1$ holds for every given term.
Q: What happens to $S_n$ as $n\to\infty$?
A: $\dfrac{1}{2\cdot3^n}\to0$, so $S_n\to n+\dfrac12$ for large $n$ — the series behaves like $n+\dfrac12$ plus a rapidly vanishing correction.
Q: Is this technique useful beyond this specific question?
A: Yes — any series of the form $\sum(c+ar^k)$, where $c$ is constant and $ar^k$ is geometric, is solved by exactly this split-and-sum method.
Prerequisites to Solve This Question
- Recognizing when a series is geometric vs. a sum of a constant and a geometric part.
- The finite geometric series summation formula.
- Algebraic manipulation of fractions with a common power-of-3 denominator.
- Verifying a closed-form answer against direct term-by-term computation.
After Solving This, You Can:
- ✔ Decompose mixed constant-geometric series before summing.
- ✔ Apply the finite GP-sum formula confidently and correctly.
- ✔ Cross-verify closed-form summation answers using small-$n$ substitution.
- ✔ Recognize and avoid the classic sign-error traps in GP-sum problems.
Study Tips for This Topic
1. Whenever a series “almost” looks geometric but isn’t quite, check if each term is a sum of a constant and a geometric piece.
2. Memorise the GP-sum formula in the form $\frac{a(1-r^n)}{1-r}$ and always verify the sign before simplifying.
3. For any closed-form summation answer, sanity-check it by plugging in $n=1$ — it must equal the first term exactly.
4. Practice recognizing numerator patterns like $3^k+1$, $2^k-1$, etc. — they appear often in JEE series questions.
Difficulty Rating & Exam Frequency
Difficulty: ★★★☆☆ (moderate)
JEE Main Frequency: Common — mixed constant-geometric series appear almost every year.
JEE Advanced Frequency: Occasional — usually as part of a longer summation problem.
Overall Importance: High for building series-manipulation fluency.
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Last Updated: August 2026
Question Source: JEE Main 2012 PYQ
Topic: Sequences And Series