Chapter 5 · Arithmetic Progressions

5.3 Sum of the First n Terms of an AP

Adding up fifty, a hundred or a thousand terms one by one is slow and easy to get wrong. Pair the first term with the last, the second with the second-last, … and every pair gives the same total — so $S_n=\tfrac n2\left[2a+(n-1)d\right]$ adds a whole AP in one line.

At the start of this chapter we walked into a cricket stadium. The front row of one stand has $20$ seats, and every row has $4$ more seats than the row in front of it. In Section 5.2 you found that the $25$th row has $20+24\times4=116$ seats. But the ground manager wants a different number: how many people can the whole stand of $25$ rows hold? That means adding $20+24+28+\dots+116$ — twenty-five numbers. You could do it on a calculator, one number at a time, and hope you never press a wrong key. Or you could use one neat idea and get the answer in a single line.

In this section we find that idea — pairing terms from the two ends — and turn it into a formula for the sum of the first $n$ terms of an AP. We use it to add up any AP, to work backwards from a sum to the number of terms (sometimes there are two answers!), to recover the terms of an AP from a formula for its sums, and to solve word problems about seats, savings, loan instalments and stacks of cans. By the end you will know exactly how many seats the stand has.

Adding an AP: the pairing trick

A well-known story is told about the mathematician Carl Friedrich Gauss. When he was about ten, his teacher asked the class to add all the whole numbers from $1$ to $100$, expecting a long, quiet hour. Young Gauss wrote down the answer almost at once. He had noticed that the numbers can be paired from the two ends:

Pair	$1+100$	$2+99$	$3+98$	$\dots$	$50+51$
Total	$101$	$101$	$101$	$\dots$	$101$

Every pair adds up to $101$, and the $100$ numbers make $50$ pairs, so the sum is $50\times101=5050$. The trick works because, as you go one step in from each end, one number goes up by $1$ while the other goes down by $1$ — the total stays the same. That is true in every AP: moving in from the front adds $d$, moving in from the back subtracts $d$. You already saw this in Section 5.2: the $k$th term from the start and the $k$th term from the end always add up to $a+l$.

📘 Definition 5.3 — The sum of the first n terms

For an AP $a_1, a_2, a_3, \dots$, the sum of the first $n$ terms is $$S_n=a_1+a_2+a_3+\dots+a_n.$$ So $S_1=a_1$, $S_2=a_1+a_2$, $S_3=a_1+a_2+a_3$, and so on. A sum of the terms of an AP written out with plus signs, such as $3+7+11+15$, is called an arithmetic series.

📐 Theorem 5.4 — Sum of the first n terms of an AP

In an AP with first term $a$ and common difference $d$, the sum of the first $n$ terms is $$S_n=\frac n2\left[2a+(n-1)d\right].$$ If the last of these $n$ terms is $l=a+(n-1)d$, this can also be written $$S_n=\frac n2\,(a+l),$$ that is, the number of terms times the average of the first and last terms.

Why it is true. Write the sum forwards, then write it again backwards underneath:

$$S_n=a+(a+d)+(a+2d)+\dots+(l-d)+l$$ $$S_n=l+(l-d)+(l-2d)+\dots+(a+d)+a$$

Now add the two lines column by column. In every column the $d$'s cancel: $a+l$, $\;(a+d)+(l-d)=a+l$, $\;(a+2d)+(l-2d)=a+l$, … There are $n$ columns, each adding up to $a+l$, so

$$2S_n=n(a+l),\qquad\text{so}\qquad S_n=\frac n2\,(a+l).$$

Finally, put $l=a+(n-1)d$ (Theorem 5.2): $\;a+l=2a+(n-1)d$, which gives $S_n=\tfrac n2\left[2a+(n-1)d\right]$. Use the first form when you know $a$ and $d$; use $\tfrac n2(a+l)$ when you know the first and last terms.

⚠️ Common mistakes

1. Writing $nd$ instead of $(n-1)d$. For $2, 5, 8, \dots$ and $n=20$: $\;S_{20}=10\left[4+19\times3\right]=610$, not $10\left[4+20\times3\right]=640$.

2. Forgetting the $2$ in $2a$. The bracket is $2a+(n-1)d$ — it is the sum of the first and last terms, $a+l$, not the last term alone.

3. Guessing $n$ in $\tfrac n2(a+l)$. If you are given the first and last terms, first count the terms with $l=a+(n-1)d$; never assume $n$ from the size of $l$.

4. Mixing up $a_n$ and $S_n$. $a_n$ is one term (the $n$th); $S_n$ is the total of the first $n$ terms. "How many seats in row $25$?" asks for $a_{25}$; "how many seats in the first $25$ rows?" asks for $S_{25}$.

✏️ Example 1 — Using the sum formula

(a) Find the sum of the first $20$ terms of $2, 5, 8, \dots$ (b) Find $41+38+35+\dots+5$. (c) Find the sum of the first $22$ terms of $8, 3, -2, \dots$

💡 Show solution▶

(a) $a=2$, $d=3$, $n=20$: $\;S_{20}=\tfrac{20}{2}\left[2\times2+19\times3\right]=10\times(4+57)=10\times61=610$.

(b) Count first $a=41$, $d=-3$, $l=5$. Number of terms: $41+(n-1)(-3)=5$ gives $(n-1)(-3)=-36$, $n-1=12$, $n=13$.

(b) Add $S_{13}=\tfrac{13}{2}(41+5)=\tfrac{13}{2}\times46=13\times23=299$.

Check In (b), pair the ends: $41+5=46$, $38+8=46$, … Six pairs make $276$, and the middle (7th) term $41-18=23$ is left over: $276+23=299$ ✓.

✏️ Example 2 — Natural numbers and odd numbers

(a) Find the sum of the first $n$ natural numbers $1+2+3+\dots+n$, and use it to find $1+2+\dots+100$ and $1+2+\dots+250$. (b) Show that the sum of the first $n$ odd numbers $1+3+5+\dots$ is $n^2$, and find $1+3+5+\dots+99$.

💡 Show solution▶

(a) Here $a=1$, $l=n$, and there are $n$ terms: $\;S_n=\tfrac n2(1+n)=\dfrac{n(n+1)}{2}$.

(a) Values $n=100$: $\;\tfrac{100\times101}{2}=5050$ (Gauss's answer). $n=250$: $\;\tfrac{250\times251}{2}=125\times251=31\,375$.

(b) $a=1$, $d=2$: $\;S_n=\tfrac n2\left[2+(n-1)\times2\right]=\tfrac n2\times2n=n^2$.

(b) Value $99$ is the $50$th odd number ($1+(n-1)\times2=99$ gives $n=50$), so $1+3+\dots+99=50^2=2500$.

Check $1+3=4=2^2$, $\;1+3+5=9=3^2$, $\;1+3+5+7=16=4^2$ ✓. The sums $1, 3, 6, 10, 15, \dots$ in (a) are called triangular numbers: they count the dots in triangles of dots with $1, 2, 3, \dots$ dots in the bottom row.

Working backwards from a sum

The formula $S_n=\tfrac n2\left[2a+(n-1)d\right]$ links five quantities: $a$, $d$, $n$, $l$ and $S_n$. Questions often give the sum and ask for something else — the number of terms, the common difference, or the terms themselves. When the unknown is $n$, the equation is usually a quadratic in $n$, which you can solve by factorisation or by the quadratic formula (Section 4.3). Remember that $n$ counts terms, so only positive whole-number roots are acceptable.

✏️ Example 3 — How many terms? Two answers!

How many terms of the AP $45, 42, 39, \dots$ must be added to give a sum of $357$? Explain why there are two answers.

💡 Show solution▶

Equation $a=45$, $d=-3$: $\;\tfrac n2\left[90+(n-1)(-3)\right]=357$, i.e. $\tfrac n2(93-3n)=357$.

Quadratic Multiply by $2$: $93n-3n^2=714$, so $3n^2-93n+714=0$, i.e. $n^2-31n+238=0$.

Solve Two numbers with product $238$ and sum $31$: $14$ and $17$. So $(n-14)(n-17)=0$ and $n=14$ or $n=17$. Both are positive whole numbers.

Why two? The terms after the $14$th are $a_{15}=3$, $a_{16}=0$, $a_{17}=-3$, and $3+0+(-3)=0$. Adding them does not change the total, so $S_{14}=S_{17}=357$. Both answers are correct.

Check $S_{14}=7\left[90+13\times(-3)\right]=7\times51=357$ ✓ and $S_{17}=\tfrac{17}{2}\left[90+16\times(-3)\right]=\tfrac{17}{2}\times42=357$ ✓.

Since $S_n=a_1+\dots+a_{n-1}+a_n$ and $S_{n-1}=a_1+\dots+a_{n-1}$, subtracting leaves just the $n$th term. This gives a way to get the terms back from a formula for the sums.

📐 Theorem 5.5 — A term from the sums

If $S_n$ is the sum of the first $n$ terms of a sequence, then $$a_1=S_1\qquad\text{and}\qquad a_n=S_n-S_{n-1}\quad\text{for } n\ge2.$$

✏️ Example 4 — The AP hidden in a sum formula

The sum of the first $n$ terms of an AP is $S_n=2n^2+3n$. Find the first term, the common difference, the $n$th term and the $12$th term.

💡 Show solution▶

First terms $a_1=S_1=2+3=5$. $\;S_2=8+6=14$, so $a_2=S_2-S_1=14-5=9$. Hence $d=9-5=4$.

nth term $a_n=a+(n-1)d=5+4(n-1)=4n+1$.

12th term $a_{12}=4\times12+1=49$.

Check By Theorem 5.5: $a_{12}=S_{12}-S_{11}=(288+36)-(242+33)=324-275=49$ ✓. A common slip is to answer $S_{12}=324$ — that is the sum of twelve terms, not the $12$th term.

✏️ Example 5 — First term, last term and sum known

The first term of an AP is $8$, the last term is $62$ and the sum of all its terms is $350$. Find the number of terms and the common difference.

💡 Show solution▶

Number of terms Use $S_n=\tfrac n2(a+l)$: $\;\tfrac n2(8+62)=350$, so $35n=350$ and $n=10$.

Common difference $l=a+(n-1)d$: $\;62=8+9d$, so $9d=54$ and $d=6$.

Check The AP is $8, 14, 20, \dots, 62$ and $S_{10}=5\left[16+9\times6\right]=5\times70=350$ ✓.

Each dot is a running total $S_n$, plotted against the number of terms $n$. Going from one dot to the next adds one more term, so the red jump is exactly $a_n=S_n-S_{n-1}$. The green line is a target sum $k$: every dot that sits on it is a solution of $S_n=k$. Press Play, or choose a situation. With a negative $d$ the sum rises, levels off and falls — so it can hit the same target twice.

Situation:$a$:45$d$:−3target $k$:$n$:14

$S_n$—

last jump $a_n=S_n-S_{n-1}$—

solutions of $S_n=k$—

Word problems: seats, savings and instalments

In a word problem the AP is hidden in a story: seats in successive rows, money saved in successive months, instalments of a loan, cans in the layers of a stack. The hardest part is deciding whether the question wants one term or a total.

🧭 Problem-solving strategy — sum problems

Write the first few terms of the AP from the story and read off $a$ and $d$. Say in words what $a_n$ stands for ("seats in row $n$", "saving in month $n$").
Decide what is asked: a single term ("in the $12$th month" → $a_n$) or a total ("in the first $12$ months altogether" → $S_n$).
Choose the formula: $\tfrac n2\left[2a+(n-1)d\right]$ if you know $a$, $d$, $n$; $\;\tfrac n2(a+l)$ if you know the first and last terms. If $n$ is unknown, expect a quadratic.
Keep only answers that make sense (a positive whole number of rows, months, prizes). For "at least" questions, take the smallest whole number that works.
Check by substituting back, and answer with units (seats, ₹, km).

✏️ Example 6 — How many seats in the stand?

A stand in a cricket stadium has $25$ rows. The front row has $20$ seats and each row has $4$ more seats than the row in front of it. (a) How many seats does the stand have? (b) The club wants to build a new stand with the same pattern ($20$ seats in front, $4$ more in each row) that holds at least $3000$ people. How many rows does it need, and how many seats are in its back row?

💡 Show solution▶

The AP Seats in rows $1, 2, 3, \dots$: $\;20, 24, 28, \dots$ with $a=20$, $d=4$. The question asks for a total, so we need $S_n$.

(a) $S_{25}=\tfrac{25}{2}\left[40+24\times4\right]=\tfrac{25}{2}\times136=25\times68=1700$ seats. (Or: $l=a_{25}=116$, and $\tfrac{25}{2}(20+116)=1700$.)

(b) Inequality $S_n=\tfrac n2\left[40+4(n-1)\right]=n(2n+18)=2n^2+18n$. We need $2n^2+18n\ge3000$, i.e. $n^2+9n-1500\ge0$.

(b) Solve The quadratic formula gives the positive root of $n^2+9n-1500=0$ as $n=\dfrac{-9+\sqrt{81+6000}}{2}=\dfrac{-9+\sqrt{6081}}{2}\approx\dfrac{-9+77.98}{2}\approx34.49$. Rows come in whole numbers, so the smallest that works is $n=35$.

(b) Check $S_{34}=2(34)^2+18(34)=2312+612=2924$ (too few) and $S_{35}=2450+630=3080$ (enough) ✓. The new stand needs $35$ rows, and its back row has $a_{35}=20+34\times4=156$ seats.

Pay-off The question from the start of the chapter is answered: the $25$-row stand holds $1700$ people — found with one multiplication instead of twenty-four additions.

Exercises

Find the sum.

1 the first $15$ terms of $3, 7, 11, \dots$
2 the first $10$ terms of $50, 46, 42, \dots$
3 the first $20$ terms of $-5, -2, 1, \dots$
4 the first $100$ terms of $0.6, 1.7, 2.8, \dots$
5 the first $12$ terms of $\tfrac16, \tfrac14, \tfrac13, \dots$
6 $5+7\tfrac12+10+\dots+65$
7 $(-4)+(-9)+(-14)+\dots+(-124)$
8 the first $10$ terms of $\sqrt2, 3\sqrt2, 5\sqrt2, \dots$

Check: type $a$, $d$ and $n$ — the sum by the formula and by adding term by term

$a$$d$$n$

last term $l$—

formula $\tfrac n2[2a+(n-1)d]$—

adding one by one—

💡 Show solutions▶

1 $a=3$, $d=4$: $\;\tfrac{15}{2}\left[6+14\times4\right]=\tfrac{15}{2}\times62=465$.

2 $a=50$, $d=-4$: $\;5\left[100+9\times(-4)\right]=5\times64=320$.

3 $a=-5$, $d=3$: $\;10\left[-10+19\times3\right]=10\times47=470$.

4 $a=0.6$, $d=1.1$: $\;50\left[1.2+99\times1.1\right]=50\times110.1=5505$.

5 $d=\tfrac14-\tfrac16=\tfrac1{12}$ (and $\tfrac13-\tfrac14=\tfrac1{12}$ ✓): $\;6\left[\tfrac13+11\times\tfrac1{12}\right]=6\times\tfrac{15}{12}=\tfrac{15}{2}$.

6 $d=2\tfrac12$. Count: $5+(n-1)\times2.5=65$ gives $n-1=24$, $n=25$. $\;S=\tfrac{25}{2}(5+65)=875$.

7 $d=-5$. Count: $-4+(n-1)(-5)=-124$ gives $n-1=24$, $n=25$. $\;S=\tfrac{25}{2}(-4-124)=\tfrac{25}{2}\times(-128)=-1600$.

8 $a=\sqrt2$, $d=2\sqrt2$: $\;5\left[2\sqrt2+9\times2\sqrt2\right]=5\times20\sqrt2=100\sqrt2$. (It is $\sqrt2$ times the sum of the first $10$ odd numbers, $10^2$.)

Find the missing quantities ($a$ = first term, $d$ = common difference, $n$ = number of terms, $l$ = last term, $S_n$ = sum).

1 $a=6$, $d=4$, $n=18$. Find $l$ and $S_n$.
2 $a=-3$, $l=87$, $n=16$. Find $d$ and $S_n$.
3 $a=12$, $d=5$, $S_n=705$. Find $n$.
4 $l=100$, $d=4$, $n=20$. Find $a$ and $S_n$.
5 $a=7$, $n=12$, $S_n=414$. Find $d$ and $l$.
6 The sum of the first $9$ terms of an AP is $171$ and its first term is $3$. Find $d$ and the $9$th term.
7 How many terms of $6, 11, 16, \dots$ must be added to give $402$?
8 How many terms of $63, 60, 57, \dots$ must be added to give $693$? Explain the double answer.

💡 Show solutions▶

1 $l=6+17\times4=74$; $\;S_{18}=9(6+74)=720$.

2 $87=-3+15d$ gives $d=6$; $\;S_{16}=8(-3+87)=672$.

3 $\tfrac n2\left[24+5(n-1)\right]=705$ gives $n(5n+19)=1410$, i.e. $5n^2+19n-1410=0$. $D=361+28\,200=28\,561=169^2$, so $n=\tfrac{-19+169}{10}=15$ (the other root is negative). $n=15$.

4 $100=a+19\times4$ gives $a=24$; $\;S_{20}=10(24+100)=1240$.

5 $6\left[14+11d\right]=414$ gives $14+11d=69$, $d=5$; $\;l=7+11\times5=62$. (Check: $6(7+62)=414$ ✓.)

6 $\tfrac92\left[6+8d\right]=171$ gives $6+8d=38$, $d=4$; $\;a_9=3+8\times4=35$. (Check: $\tfrac92(3+35)=171$ ✓.)

7 $\tfrac n2\left[12+5(n-1)\right]=402$ gives $5n^2+7n-804=0$. $D=49+16\,080=16\,129=127^2$, so $n=\tfrac{-7+127}{10}=12$. Twelve terms.

8 $\tfrac n2\left[126-3(n-1)\right]=693$ gives $n(129-3n)=1386$, i.e. $n^2-43n+462=0=(n-21)(n-22)$. So $n=21$ or $22$. Both work because $a_{22}=63+21\times(-3)=0$: adding a zero term leaves the sum unchanged.

III

Special sums, and terms from a sum formula.

1 Find the sum of the first $40$ positive integers.
2 Find the sum of the first $15$ multiples of $8$.
3 Find the sum of all odd numbers between $0$ and $50$.
4 Find the sum of all three-digit numbers that are divisible by $7$.
5 Find the sum of all two-digit numbers that leave remainder $3$ when divided by $5$.
6 The sum of the first $n$ terms of an AP is $3n^2+4n$. Find its first term, its $n$th term and its $10$th term.
7 The sum of the first $n$ terms of an AP is $9n-n^2$. Find the AP and its $15$th term.
8 Show that the sum of the first $n$ even numbers $2+4+6+\dots$ is $n(n+1)$.

💡 Show solutions▶

1 $\tfrac{40\times41}{2}=820$.

2 $8+16+\dots+120=8(1+2+\dots+15)=8\times\tfrac{15\times16}{2}=960$.

3 $1+3+\dots+49$: $49$ is the $25$th odd number, so the sum is $25^2=625$.

4 $105, 112, \dots, 994$: $\;994=105+7(n-1)$ gives $n=128$. $\;S=64(105+994)=64\times1099=70\,336$.

5 $13, 18, 23, \dots, 98$: $\;98=13+5(n-1)$ gives $n=18$. $\;S=9(13+98)=999$.

6 $a_1=S_1=7$. $\;a_n=S_n-S_{n-1}=3n^2+4n-\left[3(n-1)^2+4(n-1)\right]=3(2n-1)+4=6n+1$. $\;a_{10}=61$. (Check: $6\times1+1=7=S_1$ ✓.)

7 $a_n=S_n-S_{n-1}=9n-n^2-\left[9(n-1)-(n-1)^2\right]=9-(2n-1)=10-2n$. The AP is $8, 6, 4, 2, \dots$ ($S_1=8$ ✓), and $a_{15}=10-30=-20$.

8 $a=2$, $d=2$: $\;S_n=\tfrac n2\left[4+2(n-1)\right]=\tfrac n2(2n+2)=n(n+1)$. (Or: each even number is $2\times$ a natural number, so the sum is $2\times\tfrac{n(n+1)}{2}$.)

Applications.

1 Auditorium. An auditorium has $30$ rows of seats. The first row has $18$ seats and each row has $2$ more than the row in front. How many seats are there in all?
2 Savings jar. Meera puts ₹ $200$ in her jar in January, ₹ $250$ in February, ₹ $300$ in March, and so on. How much has she saved by the end of December?
3 Loan. A loan of ₹ $1\,17\,000$ is repaid in monthly instalments. The first instalment is ₹ $1000$ and each instalment is ₹ $200$ more than the one before. How many instalments are needed, and what is the last one?
4 Stack of cans. A shop builds a display of cans with $24$ cans in the bottom layer, $22$ in the next, $20$ in the next, and so on, with $2$ cans in the top layer. How many layers are there, and how many cans in all?
5 Training plan. Priya runs $2$ km on day $1$ and $0.25$ km more on each following day. How far does she run altogether in the first $30$ days? On which day does she first run $7$ km?
6 Prize money. A school gives ₹ $4500$ as $6$ cash prizes. Each prize is ₹ $150$ less than the one before it. Find all six prizes.
7 Clock tower. A clock strikes once at $1$ o'clock, twice at $2$ o'clock, …, twelve times at $12$ o'clock, both day and night. How many times does it strike in $24$ hours?

💡 Show solutions▶

1 $a=18$, $d=2$, $n=30$: $\;S_{30}=15\left[36+29\times2\right]=15\times94=1410$ seats.

2 $a=200$, $d=50$, $n=12$: $\;S_{12}=6\left[400+11\times50\right]=6\times950=$ ₹ $5700$.

3 $\tfrac n2\left[2000+200(n-1)\right]=117\,000$ gives $n(900+100n)=117\,000$, i.e. $n^2+9n-1170=0=(n-30)(n+39)$. So $n=30$ instalments; the last is $1000+29\times200=$ ₹ $6800$.

4 $24, 22, \dots, 2$: $\;2=24+(n-1)(-2)$ gives $n=12$ layers. $\;S_{12}=6(24+2)=156$ cans.

5 $S_{30}=15\left[4+29\times0.25\right]=15\times11.25=168.75$ km. $\;2+(n-1)\times0.25=7$ gives $n-1=20$: on day $21$.

6 First prize $a$, $d=-150$: $\;3\left[2a-750\right]=4500$ gives $2a=2250$, $a=1125$. Prizes: ₹ $1125, 975, 825, 675, 525, 375$ (they add up to $4500$ ✓).

7 In $12$ hours: $1+2+\dots+12=\tfrac{12\times13}{2}=78$ strikes. In $24$ hours: $2\times78=156$ strikes.

🎯 Exam corner Board · 1 markBoard · 2 marksBoard · 3 marksBoard · Case studyJEE Main

How this is examined: the sum formula is one of the most-tested results of the chapter — a direct sum, the sum of all multiples of a number in a range, "how many terms give this sum", $a_n$ from $S_n$, and a case study about seats, savings or practice schedules. Always write the formula first, then substitute; if you solve a quadratic for $n$, say which roots you keep and why.

E1. Board · 1 mark The sum of the first $16$ terms of the AP $10, 6, 2, \dots$ is (A) $-320$ (B) $320$ (C) $-352$ (D) $-400$

💡 Show solution▶

Idea $a=10$, $d=-4$: $\;S_{16}=8\left[20+15\times(-4)\right]=8\times(-40)=-320$.

Traps (C) comes from $8\left[20+16\times(-4)\right]$ — using $n$ instead of $n-1$. (B) loses the minus sign: most of these terms are negative, so the sum must be negative.

Answer (A) $-320$.

E2. Board · 1 mark Assertion (A): If the sum of the first $n$ terms of an AP is $S_n=n^2+2n$, then its $5$th term is $11$. Reason (R): For $n\ge2$, the $n$th term of an AP is $a_n=S_n-S_{n-1}$.
(A) Both A and R are true, and R is the correct explanation of A. (B) Both A and R are true, but R is not the correct explanation of A. (C) A is true but R is false. (D) A is false but R is true.

💡 Show solution▶

Check R $S_n$ and $S_{n-1}$ differ by exactly the $n$th term, so R is true (Theorem 5.5).

Check A $a_5=S_5-S_4=(25+10)-(16+8)=35-24=11$. A is true, and it was found using R.

Answer (A). The tempting slip is to compute $S_5=35$, call it the $5$th term, and choose (D).

E3. Board · 2 marks Find the sum of all multiples of $9$ that lie between $300$ and $700$.

💡 Show solution▶

The AP $300\div9=33.3\dots$, so the first multiple is $9\times34=306$; $\;700\div9=77.7\dots$, so the last is $9\times77=693$. AP: $306, 315, \dots, 693$, $d=9$. Number of terms: $693=306+9(n-1)$ gives $n-1=43$, $n=44$. (1 mark)

Sum $S_{44}=\tfrac{44}{2}(306+693)=22\times999=21\,978$. (1 mark)

E4. Board · 3 marks The sum of the first $6$ terms of an AP is $57$ and the sum of its first $12$ terms is $222$. Find the AP and its $20$th term.

💡 Show solution▶

Equations $S_6=3(2a+5d)=57$ gives $2a+5d=19$. $\;S_{12}=6(2a+11d)=222$ gives $2a+11d=37$. (1 mark)

Solve Subtract: $6d=18$, $d=3$; then $2a=19-15=4$, $a=2$. The AP is $2, 5, 8, 11, \dots$ (1 mark)

20th term $a_{20}=2+19\times3=59$. (1 mark) Check: $S_6=3(4+15)=57$ ✓.

E5. Board · Case study A young fast bowler at a cricket academy follows a practice plan. On day $1$ she bowls $40$ balls, and each day she bowls $5$ balls more than the day before.
(i) How many balls does she bowl on day $15$? (1 mark)
(ii) How many balls does she bowl altogether in the first $10$ days? (1 mark)
(iii) After how many days will she have bowled $1240$ balls in total? (2 marks)

💡 Show solution▶

The AP Balls per day: $40, 45, 50, \dots$ with $a=40$, $d=5$.

(i) One day's balls is a term: $a_{15}=40+14\times5=110$ balls.

(ii) A total is a sum: $S_{10}=5\left[80+9\times5\right]=5\times125=625$ balls.

(iii) $\tfrac n2\left[80+5(n-1)\right]=1240$ gives $n(5n+75)=2480$, i.e. $n^2+15n-496=0$. Factorise: $(n-16)(n+31)=0$. Since $n\gt0$, $n=16$: after $16$ days. Check: $S_{16}=8\left[80+75\right]=1240$ ✓.

E6. JEE Main The sums of the first $n$ terms of two APs are in the ratio $(3n+1):(2n+5)$ for every $n$. The ratio of their $8$th terms is (A) $46:35$ (B) $25:21$ (C) $4:3$ (D) $23:19$

💡 Show solution▶

Key fact For an odd number of terms, $S_{2m-1}=\tfrac{2m-1}{2}(a+l)=(2m-1)\times(\text{middle term})$, and the middle term is the $m$th. So the $8$th term is $S_{15}\div15$.

Ratio $\dfrac{a_8}{b_8}=\dfrac{S_{15}/15}{S'_{15}/15}=\dfrac{S_{15}}{S'_{15}}=\dfrac{3\times15+1}{2\times15+5}=\dfrac{46}{35}$.

Check Take $S_n=n(3n+1)$ and $S'_n=n(2n+5)$. By Theorem 5.5 the terms are $a_n=6n-2$ and $b_n=4n+3$, so $a_8:b_8=46:35$ ✓.

Answer (A). (B) comes from putting $n=8$ into the ratio of the sums — that compares $S_8$ with $S'_8$, not the $8$th terms.

🔬 Research–Teaching NexusMini project · optional, about 45 minutes

Research question. A popular "52-week savings challenge" asks you to save ₹ $10$ in week $1$, ₹ $20$ in week $2$, ₹ $30$ in week $3$, and so on for a year. How much does it really add up to — and is it a better plan than saving the same fixed amount every week?

Why it matters. Banks and money-education programmes in many countries promote challenges like this because small, steadily growing amounts are easier to start than one big saving. But people often misjudge the total: a growing plan feels small at first and becomes hard near the end. Knowing the sum of an AP lets you see the whole year at once before you begin.

What to do.

Write the weekly savings as an AP. What are $a$ and $d$? Use $S_n$ to find the total saved after $52$ weeks.
Make a spreadsheet with columns "week $n$", "saving $a_n$" and "total so far $S_n$" (each total is the previous total plus this week's saving). Check that row $52$ agrees with your formula.
Your family's goal is ₹ $5000$. Use $S_n\ge5000$ to find the first week in which the total reaches the goal, then confirm it in the spreadsheet.
Compare with a "flat" plan that saves the same amount every week and reaches the same yearly total. How much is that each week? Which plan is easier in January, and which in December?
Ask three adults which plan they would choose and why. Try a "reverse" challenge (₹ $520$ in week $1$, ₹ $10$ less each week). What is its total, and why might it suit some savers better?

Tools: Phone or computer spreadsheet, notebook, a short family survey

Hand in: Your AP and the yearly total by formula, the spreadsheet screenshot, the week in which you reach ₹ $5000$, the comparison with the flat plan, and two sentences on which plan you would follow.

Going further. Banks pay interest on savings. Find out roughly what a savings account pays in a year and estimate how much extra the challenge would earn. (Interest that grows on interest does not form an AP — you will meet that kind of growth in Class 11.)

Teacher/parent note: $a=d=10$, so $S_n=5n(n+1)$ and $S_{52}=$ ₹ $13\,780$. The goal of ₹ $5000$ needs $n(n+1)\ge1000$: $S_{31}=4960$, $S_{32}=5280$, so week $32$. The flat plan is $13\,780\div52=$ ₹ $265$ a week. The reverse challenge has the same terms in the opposite order, so the same total ₹ $13\,780$ — a nice check on the pairing idea.

🧪 Self-check quiz

0/5 correct

Q1. The sum of the first $10$ terms of $1, 3, 5, \dots$ is

Q2. $5+10+15+\dots+100=$

Q3. The sum of the first $n$ terms of an AP is $S_n=n^2+n$. Its $3$rd term is

Q4. How many terms of $1, 2, 3, \dots$ add up to $210$?

Q5. For the AP $21, 18, 15, \dots$, how many terms must be added to get a sum of $0$?

← 5.2 The nth Term of an AP Chapter 6 — Triangles →