If the sum of the first four terms of an A.P. is 6 and the sum of its first six terms is 4, then the sum of its first twelve terms is
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Solution Steps
1
Write given sums using AP formula
Let first term = $a$, common difference = $d$.
$S_n = \frac{n}{2}[2a + (n-1)d]$.
$S_4 = \frac{4}{2}[2a + 3d] = 2(2a+3d) = 6$ ⟹ $2a+3d = 3$ …(1)
$S_6 = \frac{6}{2}[2a + 5d] = 3(2a+5d) = 4$ ⟹ $2a+5d = \frac{4}{3}$ …(2)
2
Solve for a and d
Subtract (1) from (2): $(2a+5d) – (2a+3d) = \frac{4}{3} – 3$
$2d = \frac{4}{3} – \frac{9}{3} = -\frac{5}{3}$ ⟹ $d = -\frac{5}{6}$.
From (1): $2a + 3\left(-\frac{5}{6}\right) = 3$
$2a – \frac{5}{2} = 3$ ⟹ $2a = 3 + \frac{5}{2} = \frac{11}{2}$ ⟹ $a = \frac{11}{4}$.
3
Compute $S_{12}$
$S_{12} = \frac{12}{2}[2a + 11d] = 6\left[2\cdot\frac{11}{4} + 11\left(-\frac{5}{6}\right)\right]$
$= 6\left[\frac{11}{2} – \frac{55}{6}\right] = 6\left[\frac{33}{6} – \frac{55}{6}\right] = 6 \times \frac{-22}{6} = -22$.
4
Verify using alternative method ($S_n = An^2 + Bn$)
Assume $S_n = An^2 + Bn$. Then $S_4 = 16A + 4B = 6$, $S_6 = 36A + 6B = 4$.
Divide first by 2: $8A + 2B = 3$. Second divide by 2: $18A + 3B = 2$.
Solve: multiply first by 3: $24A + 6B = 9$; second by 2: $36A + 6B = 4$. Subtract: $-12A = 5$ ⇒ $A = -5/12$. Then $8(-5/12)+2B=3$ ⇒ $-10/3 + 2B = 3$ ⇒ $2B = 19/3$ ⇒ $B = 19/6$.
$S_{12} = A\cdot144 + B\cdot12 = 144(-5/12) + 12(19/6) = -60 + 38 = -22$ ✓.
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Key Insight
Even without finding a and d explicitly, using $S_n = An^2 + Bn$ gives a quick route to $S_{12}$.
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Arithmetic Progression – Theory (expanded)
1. General form and nth term
An arithmetic progression (AP) is a sequence where the difference between consecutive terms is constant. If the first term is $a$ and common difference $d$, then the $n^{th}$ term is $T_n = a + (n-1)d$. This linear nature makes sums behave quadratically. The behaviour of the sequence (increasing if $d>0$, decreasing if $d<0$) is determined entirely by $d$.
2. Sum of first n terms – standard formula
$S_n = \frac{n}{2}[2a + (n-1)d] = \frac{n}{2}(a + T_n)$. This formula is derived by pairing terms equidistant from ends. It is quadratic in $n$ because it can be written as $S_n = \frac{d}{2}n^2 + (a – \frac{d}{2})n$. Therefore $S_n$ is a quadratic function of $n$, which is a powerful observation used in the alternative solution.
3. Alternative representation $S_n = An^2 + Bn$
Since $S_n$ is always a quadratic in $n$ (with no constant term because $S_0 = 0$), we can set $S_n = An^2 + Bn$. Then $A = d/2$ and $B = a – d/2$. This method is especially useful when we are given $S_p$ and $S_q$ and need $S_r$ without explicitly computing $a$ and $d$. It reduces the problem to solving two linear equations in $A$ and $B$, then plugging $n=r$. This technique is frequently tested in JEE.
4. Properties of sums in AP
For an AP, $S_n – S_{n-1} = T_n$. Also, the sequence of successive differences of sums, i.e., $S_1, S_2 – S_1, S_3 – S_2, …$ forms the original AP. Moreover, $S_{k}, S_{2k} – S_{k}, S_{3k} – S_{2k}$ are also in AP. Such properties help in quick checks. In this problem, note that $S_4=6$, $S_6=4$ implies that $T_5+T_6 = S_6 – S_4 = -2$, which is consistent with the negative $d$ we found.
5. Why $S_{12}$ is negative despite small positive $S_4$
Even though first four terms sum to 6 (positive), the common difference $d = -5/6$ is negative, so after a certain number of terms, the terms themselves become negative and eventually the cumulative sum starts decreasing. By the 12th term, the sum has dipped to –22. This illustrates the importance of sign of $d$ in long‑term behaviour of the AP.
6. Quick verification using $T_n$
We can also compute the terms: $a = 11/4 = 2.75$, $d = -0.8333$. Then $T_1=2.75$, $T_2≈1.9167$, $T_3≈1.0833$, $T_4≈0.25$, sum first 4 ≈ 6.0. $T_5≈-0.5833$, $T_6≈-1.4167$, sum first 6 ≈ 4.0 (correct). Continue: $T_7≈-2.25$, $T_8≈-3.083$, $T_9≈-3.917$, $T_{10}≈-4.75$, $T_{11}≈-5.583$, $T_{12}≈-6.416$. Summing terms 7‑12 ≈ –25.999, plus first 6 sum 4 gives –22.0. Consistent.
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FAQs – Arithmetic Progression
Q
What does ‘sum of first n terms’ mean?
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It means Sₙ = T₁ + T₂ + … + Tₙ, where Tₖ is the k‑th term of the AP.
Q
Why is Sₙ a quadratic in n?
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Because Sₙ = n/2[2a+(n-1)d] = (d/2)n² + (a – d/2)n, which is of form An²+Bn.
Q
Can we find S₁₂ without finding a and d?
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Yes, set Sₙ = An²+Bn, use given S₄ and S₆ to solve A,B, then compute S₁₂ = 144A+12B.
Q
What is the common difference in this problem?
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d = –5/6 ≈ –0.8333, found by solving the equations.
Q
Why is S₁₂ negative while S₄ positive?
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Because d is negative, after some terms the cumulative sum decreases and eventually becomes negative.
Q
What if S₆ were larger than S₄?
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That would imply d positive, making S₁₂ larger positive. But here S₆ < S₄, confirming d<0.
Q
What is the formula for Sₙ in terms of last term?
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Sₙ = n/2 (first term + last term). Here last term T₁₂ = a+11d = 11/4 + 11(-5/6) = –77/12, so S₁₂ = 12/2 (11/4 – 77/12) = 6(33/12 – 77/12) = 6(-44/12) = –22.
Q
How can I check if my a and d are correct?
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Plug them back into S₄ and S₆; you should get 6 and 4 respectively.
Q
What is the 12th term of this AP?
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T₁₂ = a + 11d = 11/4 + 11(–5/6) = 11/4 – 55/6 = (33 – 110)/12 = –77/12 ≈ –6.4167.
Q
Is there a direct relation between S₄, S₆ and S₁₂?
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From quadratic fit, S₁₂ = 3S₆ – 3S₄ + S₀? Not linear. Better to use A,B method.
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