1. Finding the (n^{th}) Term

What is the question?

These questions usually ask:

“Find the 10th term of the AP.”

Or:

“Which term of the AP is equal to ___?”

Idea

In an AP, the difference between consecutive terms is always the same.

3, 7, 11, 15, 19, ...

    +4  +4  +4  +4

The important formula is:

$$a_n=a+(n-1)d$$

where:

a = first term
d = common difference
n = term number
aₙ = nth term

Example

Question: Find the 10th term of:

3, 7, 11, 15, ...

Step 1: Identify (a) and (d)

a = 3

d = 7 - 3
  = 4

Step 2: Use the formula

$$a_{10}=a+(10-1)d$$ $$=3+9(4)$$ $$=39$$

Answer: 39

Remember

Nth term = first term + (n − 1) × common difference

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2. Finding the Number of Terms

What is the question?

These questions usually ask:

“How many terms are there in the AP?”

Or:

“The first term and last term are given. Find the number of terms.”

Idea

Use the nth-term formula:

$$l=a+(n-1)d$$

where (l) is the last term.

Rearranging:

$$n=\frac{l-a}{d}+1$$

Example

Question: How many terms are there in:

4, 8, 12, ..., 40

Step 1: Identify the values

a = 4
d = 4
l = 40

Step 2: Use the formula

$$n=\frac{40-4}{4}+1$$ $$n=9+1$$ $$n=10$$

Answer: 10 terms

Remember

Number of terms
= (Last - First) / Difference + 1

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3. Sum of the First (n) Terms

What is the question?

These questions ask:

“Find the sum of the first 10 terms.”

Or:

“Find the sum of an AP up to a particular term.”

Idea

Once you know the first term, common difference and number of terms, use:

$$S_n=\frac n2[2a+(n-1)d]$$

If the last term is known, an easier formula is:

$$S_n=\frac n2(a+l)$$

Example

Question: Find the sum of the first 10 terms of:

3, 7, 11, 15, ...

We already know:

a = 3
d = 4
n = 10

Use:

$$S_{10}=\frac{10}{2}[2(3)+(10-1)(4)]$$ $$=5[6+36]$$ $$=5(42)$$ $$=210$$

Answer: 210

Shortcut

If you know the last term:

$$S_n=\frac n2(a+l)$$

For example, the 10th term is 39:

$$S_{10}=\frac{10}{2}(3+39)$$ $$=5(42)=210$$

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4. Finding a Term from the Sum

What is the question?

These questions give the sum of terms and ask you to find a particular term or the AP.

For example:

“The sum of the first (n) terms is given. Find the (n^{th}) term.”

Idea

Use:

$$S_n=\frac n2[2a+(n-1)d]$$

Then use the information given to find (a), (d), or (n).

Example

Question: The sum of the first 10 terms of an AP is 150 and the first term is 6. Find the common difference.

Use:

$$150=\frac{10}{2}[2(6)+9d]$$ $$150=5(12+9d)$$ $$30=12+9d$$ $$9d=18$$ $$d=2$$

Answer: d = 2

Remember

If the question gives a sum, start with the (S_n) formula.

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5. Equidistant Terms Property

What is the question?

These questions ask things like:

“In an AP, find the sum of the 6th and 10th terms.”

Or:

“Find two terms whose positions are equally far from the beginning and end.”

Idea

In an AP, terms equally far from the two ends have the same sum.

For an AP with (n) terms:

$$a_k+a_{n-k+1}=a_1+a_n$$

So:

1st + last
=
2nd + second-last
=
3rd + third-last
=
...

Example

Question: An AP has 15 terms. The first term is 5 and the last term is 47. Find the sum of the 6th and 10th terms.

Notice:

6th + 10th

6 + 10 = 16

Since there are 15 terms:

15 - 6 + 1 = 10

So they are equidistant from the ends.

Therefore:

$$a_6+a_{10}=a_1+a_{15}$$ $$=5+47$$ $$=52$$

Answer: 52

Visual

1   2   3   4   5   6 ... 10  11  12  13  14  15
|                                       |
5                                      47

6th -------------------------- 10th

6th + 10th = 1st + 15th

Shortcut

Equidistant terms → use first term + last term.

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6. Middle Term of an AP

What is the question?

These questions ask:

“Find the middle term of an AP.”

Or:

“Find the two middle terms.”

Idea

It depends on whether the number of terms is odd or even.

Odd number of terms:
One middle term

Even number of terms:
Two middle terms

For (n) terms:

Odd:
Middle term = (n + 1)/2

Even:
Middle terms = n/2 and (n/2 + 1)

Example

Question: Find the middle term(s) of:

4, 8, 12, ..., 40

First find the number of terms:

$$n=\frac{40-4}{4}+1=10$$

There are 10 terms, so there are two middle terms:

10/2 = 5th

5 + 1 = 6th

Find them:

$$a_5=4+(5-1)(4)=20$$ $$a_6=4+(6-1)(4)=24$$

Answer: 20 and 24

Useful shortcut

For an AP with an even number of terms:

$$\text{Average of the two middle terms} ====================================== \frac{a+l}{2}$$

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7. Finding the AP from Two Given Terms

What is the question?

These questions give two terms of an AP and ask:

“Find the first term and common difference.”

For example:

“The 5th term is 20 and the 9th term is 40. Find the AP.”

Idea

Write each term using:

$$a_n=a+(n-1)d$$

Then solve the two equations.

Example

Question: The 5th term of an AP is 20 and the 9th term is 40. Find the AP.

Step 1: Write the 5th term

$$a+4d=20$$

Step 2: Write the 9th term

$$a+8d=40$$

Subtract:

$$4d=20$$ $$d=5$$

Now:

$$a+4(5)=20$$ $$a=0$$

Therefore:

a = 0
d = 5

The AP is:

0, 5, 10, 15, 20, ...

Remember

Two known terms → make two equations → solve for (a) and (d).

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8. Combined AP Equations

What is the question?

These are slightly harder questions where they give relationships such as:

“The sum of the 5th and 9th terms is 40, and the sum of the 8th and 14th terms is 64. Find the AP.”

Idea

Write every term using:

$$a_n=a+(n-1)d$$

Then create two equations.

Example

Question: The sum of the 5th and 9th terms is 40, and the sum of the 8th and 14th terms is 64. Find the AP.

Step 1: Write the first condition

5th term:

$$a+4d$$

9th term:

$$a+8d$$

Their sum is 40:

$$(a+4d)+(a+8d)=40$$ $$2a+12d=40$$

Divide by 2:

$$a+6d=20$$

Step 2: Write the second condition

8th term:

$$a+7d$$

14th term:

$$a+13d$$

Their sum is 64:

$$(a+7d)+(a+13d)=64$$ $$2a+20d=64$$

Divide by 2:

$$a+10d=32$$

Step 3: Subtract

a + 10d = 32
a +  6d = 20
----------------
     4d = 12

d = 3

Now:

$$a+6(3)=20$$ $$a=2$$

Therefore:

a = 2
d = 3

The AP is:

2, 5, 8, 11, 14, ...

Answer: 2, 5, 8, 11, ...

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Advanced AP Variants

These are less direct but can appear in tougher aptitude questions. Don’t delete them from your preparation; just study them after the basic patterns.

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9. Three Terms in AP

What is the question?

You may be given three numbers that are in AP and asked to find an unknown.

Idea

If three numbers are in AP:

$$2b=a+c$$

An extremely useful way to represent three AP terms is:

$$a-d,\quad a,\quad a+d$$

Example

Question: Three numbers are in AP. Their sum is 30 and their product is 960. Find the numbers.

Since their sum is 30:

$$3a=30$$ $$a=10$$

So write the numbers as:

10-d, 10, 10+d

Their product is 960:

$$(10-d)(10)(10+d)=960$$ $$10(100-d^2)=960$$ $$100-d^2=96$$ $$d^2=4$$ $$d=2$$

Therefore:

8, 10, 12

Answer: 8, 10, 12

Remember

Three AP terms → (a-d,\ a,\ a+d) is often the fastest setup.

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10. Finding a Term Using Another Term

What is the question?

For example:

“If the 5th term is 18 and the 12th term is 46, find the 20th term.”

Idea

First find (d), then find (a), then find the required term.

Example

Question: The 5th term of an AP is 18 and the 12th term is 46. Find the 20th term.

Write:

$$a+4d=18$$ $$a+11d=46$$

Subtract:

$$7d=28$$ $$d=4$$

Then:

$$a+16=18$$ $$a=2$$

Now:

$$a_{20}=2+19(4)$$ $$=78$$

Answer: 78