Use T₅ and T₁₀ as two equations to find first term a and common difference d.

The 5th Term is 28 and 10th Term is 53 — Find the AP

Question

The 5th term of an AP is 28 and the 10th term is 53. Find the AP.

Solution — Step by Step

The general term of an AP is $T_n = a + (n-1)d$ , where $a$ is the first term and $d$ is the common difference.

Plugging in the given information:

T_5 = a + 4d = 28 \quad \text{...(i)}

T_{10} = a + 9d = 53 \quad \text{...(ii)}

We subtract equation (i) from equation (ii). The reason: both equations have $a$ , so subtraction cancels it out and leaves only $d$ .

(a + 9d) - (a + 4d) = 53 - 28

5d = 25

\boxed{d = 5}

Put $d = 5$ into equation (i):

a + 4(5) = 28

a + 20 = 28

\boxed{a = 8}

With $a = 8$ and $d = 5$ , the AP is:

8,\ 13,\ 18,\ 23,\ 28,\ 33,\ 38,\ 43,\ 48,\ 53,\ \ldots

We can verify: the 5th term is $8 + 4(5) = 28$ ✓ and the 10th term is $8 + 9(5) = 53$ ✓

Why This Works

Every AP is completely determined by just two values: the first term $a$ and the common difference $d$ . Once we know these, we know every term. So two conditions (here, $T_5$ and $T_{10}$ ) give us two equations — exactly enough to solve for two unknowns.

The clever part of Step 2 is choosing to subtract rather than substitute directly. Subtraction eliminates $a$ in one clean move, which is faster than isolating $a$ from one equation and substituting into the other. Both approaches give the same answer — the subtraction method just saves a line of algebra.

This is a standard simultaneous equations setup dressed in AP language. Once you recognise that $T_n = a + (n-1)d$ turns any term condition into a linear equation in $a$ and $d$ , the rest is Class 10 algebra.

Alternative Method

We can use the gap between the two terms directly. The 10th term comes 5 terms after the 5th term, so it’s exactly $5d$ more.

T_{10} - T_5 = 5d

53 - 28 = 5d

d = 5

This is the same arithmetic but framed differently — you’re using the fact that going from the 5th to the 10th term means adding $d$ exactly 5 times. Then find $a$ from $T_5 = a + 4d = 28$ as before.

When two terms $T_m$ and $T_n$ are given, the common difference is always:

d = \frac{T_n - T_m}{n - m}

This shortcut saves setup time in MCQ sections. For this problem: $d = \frac{53 - 28}{10 - 5} = \frac{25}{5} = 5$ .

Common Mistake

The most frequent error here is writing $T_5 = a + 5d$ instead of $a + 4d$ . Students forget that the formula is $a + (n-1)d$ , not $a + nd$ .

Think of it this way: the 1st term is $a + 0d$ (zero gaps from the start), the 2nd term is $a + 1d$ (one gap), and so on. The 5th term has 4 gaps from the first term, giving $a + 4d$ .

If you use $a + 5d = 28$ and $a + 10d = 53$ , you’ll still get $d = 5$ , but your $a$ will be wrong ( $a = 3$ instead of $8$ ), and the entire AP shifts by one term.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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