nth term = a + (n-1)d. Here a=3, d=4, n=20. So T₂₀ = 3 + 19×4 = 79.

Find the 20th Term of AP: 3, 7, 11, 15... — Using nth Term Formula

Question

Find the 20th term of the arithmetic progression: 3, 7, 11, 15, …

Solution — Step by Step

The first term is $a = 3$ . The common difference $d = 7 - 3 = 4$ (subtract any consecutive pair — same result throughout the AP).

The general term of any AP is:

T_n = a + (n-1)d

This formula comes directly from the idea that to reach the $n$ th term, we start at $a$ and add $d$ exactly $(n-1)$ times — not $n$ times. That off-by-one is where most marks are lost.

T_{20} = 3 + (20-1) \times 4

T_{20} = 3 + 19 \times 4

T_{20} = 3 + 76 = \mathbf{79}

We know $T_1 = 3$ , $T_2 = 7$ , $T_3 = 11$ . The pattern holds: each term is 4 more than the previous. At $T_{20}$ , we’ve added $d$ a total of 19 times from $T_1$ , so $3 + 76 = 79$ checks out.

Why This Works

An AP is just a sequence where each gap is constant. That constant gap is $d$ . So the 2nd term is $a + d$ , the 3rd is $a + 2d$ , and the $n$ th term is $a + (n-1)d$ — because we take $d$ exactly $(n-1)$ steps to get from the 1st to the $n$ th term.

Think of it like a staircase: if each step is 4 cm high and you’re on the 1st step at height 3 cm, then step 20 is at $3 + 19 \times 4 = 79$ cm. The number of gaps between 20 steps is 19, not 20.

This formula is fundamental — NCERT Class 10 Chapter 5 builds everything else (sum of AP, finding number of terms, middle term) on top of this single idea.

Alternative Method

We can use the recursive approach: write out the pattern.

T_n = T_{n-1} + 4

From $T_1 = 3$ , the $n$ th term is $3 + (n-1) \times 4$ .

This is slower than the formula but useful for verifying your answer or when you’ve forgotten the formula mid-exam. Count: $T_5 = 19$ , $T_{10} = 39$ , $T_{15} = 59$ , $T_{20} = 79$ . Each jump of 5 terms adds $5 \times 4 = 20$ to the value.

In board exams, this “jumping by 5 terms” trick saves time for verification. Jump from $T_{15} = 59$ to $T_{20}$ : add $5 \times 4 = 20$ , giving 79. Done in 10 seconds.

Common Mistake

The most common error: using $n \times d$ instead of $(n-1) \times d$ .

Students write $T_{20} = 3 + 20 \times 4 = 83$ . This is wrong — it gives you a term that doesn’t exist in the sequence.

Remember: between the 1st and 20th term, there are only 19 gaps, not 20. The formula is $a + (n-1)d$ , and that $(n-1)$ is non-negotiable. This exact slip costs 2-3 marks every year in CBSE board papers.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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