If S₂₀ = 30 and S₃₀ = 20 for an Arithmetic Progression, Find S₅₀

A standard arithmetic progression problem, not tied to a specific exam paper.

Problem

If $S_{20}=30$ and $S_{30}=20$ for an arithmetic progression, find $S_{50}$.

Key insight. The two given sums are just two linear equations in the AP’s first term $a$ and common difference $d$ — solving that pair pins down both unknowns completely, after which $S_{50}$ is a direct substitution.

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Approach

Write both $S_{20}$ and $S_{30}$ using the standard AP sum formula $S_n=\dfrac{n}2\left[2a+(n-1)d\right]$, giving two linear equations in $a$ and $d$. Solve them simultaneously, then substitute the resulting values of $a$ and $d$ into the formula for $S_{50}$.

Solution

Step 1 — Write S20 in terms of a and d

$$S_{20} = \frac{20}2\left[2a+19d\right] = 10(2a+19d) = 30$$

$$2a+19d = 3 \quad \text{…(i)}$$

Step 2 — Write S30 in terms of a and d

$$S_{30} = \frac{30}2\left[2a+29d\right] = 15(2a+29d) = 20$$

$$2a+29d = \frac{20}{15} = \frac43 \quad \text{…(ii)}$$

Step 3 — Solve for d by subtracting

Subtracting (i) from (ii) eliminates $2a$, since it appears identically in both equations:

$$(29d-19d) = \frac43-3 = \frac43-\frac93 = -\frac53$$

$$10d = -\frac53 \implies d = -\frac1{6}$$

Step 4 — Solve for a

Substituting $d=-\dfrac16$ back into equation (i):

$$2a = 3-19\left(-\frac16\right) = 3+\frac{19}6 = \frac{18}6+\frac{19}6 = \frac{37}6$$

So $2a=\dfrac{37}6$ — this combination is exactly what’s needed for the next step, so there’s no need to divide further to isolate $a$ itself.

Step 5 — Compute S50

$$S_{50} = \frac{50}2\left[2a+49d\right] = 25\left[2a+49d\right]$$

With $2a=\dfrac{37}6$ and $49d=49\times\left(-\dfrac16\right)=-\dfrac{49}6$:

$$2a+49d = \frac{37}6-\frac{49}6 = -\frac{12}6 = -2$$

$$S_{50} = 25\times(-2) = -50$$

Answer

Common mistakes

• Sign errors when subtracting the two equations. Since the answer involves a negative common difference, it’s easy to lose a minus sign while isolating $d$ — always double-check the subtraction by substituting $d$ back into both original equations.
• Dividing by 2 unnecessarily when solving for $2a$. Since only the combination $2a+49d$ is needed at the end, solving explicitly for $a$ (rather than keeping $2a$ as a single quantity) just adds an extra place to make an arithmetic slip.
• Forgetting that a negative sum is a perfectly valid answer. Because $d$ is negative here, the terms of the AP eventually turn negative and large enough to make $S_{50}$ negative — there’s no algebraic reason to expect a positive result.

Practise next

• If $S_{10}=100$ and $S_{20}=100$ for an AP, find $S_{30}$, using the same pair-of-equations approach — and notice what the equal sums tell you about $d$ before you even solve.