Arithmetic Progressions

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CLASS X Mathematics ~5 marks/year Ch 5 of 14
Arithmetic Progressions

Class 10 · Mathematics · NCERT chapter notes · Akanksha Classes

Snapshot
  • An Arithmetic Progression (AP) is a list of numbers where each term is got by adding a fixed number $d$ (the common difference) to the previous one.
  • Two pieces of information fix an entire AP: the first term $a$ and the common difference $d$. General form: $a,\ a+d,\ a+2d,\ a+3d,\dots$
  • nth term: $a_n=a+(n-1)d$ — jump straight to any term without listing them all.
  • Sum of first $n$ terms: $S_n=\dfrac{n}{2}[2a+(n-1)d]=\dfrac{n}{2}(a+a_n)$ — Gauss's pairing trick.
  • Board weightage: ~5 marks/year — usually one "find the nth term / which term" question and one "sum of n terms" word problem (logs, prizes, savings, salaries).
Detailed notes

1. Patterns around us — why this chapter exists

Nature and daily life are full of patterns: petals of a sunflower, holes of a honeycomb, spirals on a pine cone. NCERT opens with everyday lists:

  • A salary of $\rupee$8000 rising by $\rupee$500 a year: $8000,\ 8500,\ 9000,\dots$
  • Ladder-rung lengths dropping by 2 cm: $45,\ 43,\ 41,\ 39,\dots$
  • Savings of $\rupee$100 increased by $\rupee$50 each birthday: $100,\ 150,\ 200,\ 250,\dots$

In each, the next term comes from the previous by adding a fixed number. Lists like this are Arithmetic Progressions. Some other lists (areas of squares $1^2,2^2,3^2,\dots$ or the rabbit/Fibonacci list $1,1,2,3,5,8,\dots$) are not APs — they grow by a different rule. This chapter studies only the "add a fixed number" pattern, learns to find its $n$th term and the sum of $n$ terms, and uses these in word problems.

2. Definition of an AP and the common difference

An arithmetic progression is a list of numbers in which each term is obtained by adding a fixed number to the preceding term — except the first term. That fixed number is the common difference, written $d$. It can be positive, negative or zero.

Denote the terms $a_1,a_2,a_3,\dots,a_n$. Then by definition:

$$a_2-a_1=a_3-a_2=\dots=a_n-a_{n-1}=d$$

Look at NCERT's sample lists:

  • $1,2,3,4,\dots$ — each term is $1$ more than before, so $d=1$.
  • $100,70,40,10,\dots$ — each is $30$ less, so $d=-30$.
  • $-3,-2,-1,0,\dots$ — add $1$ each time, $d=1$.
  • $3,3,3,3,\dots$ — add $0$ each time, $d=0$ (a constant AP).
  • $-1.0,-1.5,-2.0,-2.5,\dots$ — subtract $0.5$, so $d=-0.5$.

Caution: to find $d$, always subtract a term from the term that follows it: $d=a_{k+1}-a_k$. For $6,3,0,-3,\dots$ we compute $3-6=-3$, not $6-3$, even though the later term is smaller.

3. General form, finite vs infinite, and how to test an AP

If the first term is $a$ and the common difference is $d$, the AP is:

$$a,\ a+d,\ a+2d,\ a+3d,\ \dots$$

This is the general form. So to know an AP completely you need both $a$ and $d$. For example $a=6,d=3$ gives $6,9,12,15,\dots$ while $a=6,d=-3$ gives $6,3,0,-3,\dots$

  • An AP with a last term is finite (e.g. heights $147,148,\dots,157$).
  • An AP that never ends is infinite (e.g. $1,2,3,4,\dots$).

To test whether a list is an AP: compute consecutive differences $a_2-a_1,\ a_3-a_2,\ a_4-a_3,\dots$ If they are all equal, it is an AP; otherwise it is not. (You only need to confirm the difference is the same throughout — one mismatch disqualifies it.) The list $1,1,2,3,5,\dots$ fails this test, so it is not an AP.

4. First worked examples on identifying an AP

NCERT Example 1 — write $a$ and $d$

For the AP $\dfrac{3}{2},\ \dfrac{1}{2},\ -\dfrac{1}{2},\ -\dfrac{3}{2},\dots$ the first term is $a=\dfrac{3}{2}$. The common difference is $d=\dfrac{1}{2}-\dfrac{3}{2}=-1$. (Any two consecutive terms give the same $d$.)

NCERT Example 2 — is it an AP? give next two terms

(i) $4,10,16,22,\dots$: differences $10-4=6,\ 16-10=6,\ 22-16=6$ — equal. So it is an AP with $d=6$. Next two terms: $22+6=28$ and $28+6=34$.

(ii) $1,-1,-3,-5,\dots$: differences $-1-1=-2,\ -3-(-1)=-2,\ -5-(-3)=-2$ — equal. AP with $d=-2$. Next two: $-5+(-2)=-7$ and $-7+(-2)=-9$.

(iii) $-2,2,-2,2,-2,\dots$: $a_2-a_1=2-(-2)=4$ but $a_3-a_2=-2-2=-4$. Differences differ, so not an AP.

(iv) $1,1,1,2,2,2,3,3,3,\dots$: $a_2-a_1=0$ but $a_4-a_3=2-1=1$. Not equal, so not an AP.

5. The nth term of an AP

Suppose Reena's salary starts at $\rupee$8000 and rises by $\rupee$500 a year. The pattern is:

$$a_1=a,\quad a_2=a+d,\quad a_3=a+2d,\quad a_4=a+3d,\ \dots$$

The coefficient of $d$ is always one less than the term number. So the $n$th term (also called the general term) is:

$$\boxed{\,a_n=a+(n-1)d\,}$$

If the AP is finite with $m$ terms, the last term $a_m$ is often written $l$, so $l=a+(m-1)d$. This one formula handles "find a given term", "which term equals…", and "is this number in the AP?" type questions.

NCERT Example 3 — 10th term

AP $2,7,12,\dots$ so $a=2,\ d=7-2=5,\ n=10$. Then $a_{10}=a+(10-1)d=2+9\times5=2+45=47$. The 10th term is $\mathbf{47}$.

NCERT Example 4 — which term is $-81$? is any term $0$?

AP $21,18,15,\dots$ so $a=21,\ d=-3$. Set $a_n=-81$: $-81=21+(n-1)(-3)\Rightarrow-81=24-3n\Rightarrow-105=-3n\Rightarrow n=35$. So the 35th term is $-81$.

For $a_n=0$: $21+(n-1)(-3)=0\Rightarrow 3(n-1)=21\Rightarrow n-1=7\Rightarrow n=8$. So the 8th term is $0$.

NCERT Example 5 — find the AP from two terms

3rd term is $5$ and 7th term is $9$. So $a+2d=5$ …(1) and $a+6d=9$ …(2). Subtracting (1) from (2): $4d=4\Rightarrow d=1$, then $a=5-2(1)=3$. The AP is $3,4,5,6,7,\dots$

6. More nth-term examples — membership and last-term tricks

NCERT Example 6 — is $301$ a term of $5,11,17,23,\dots$?

Differences are all $6$, so it is an AP with $a=5,\ d=6$. Let $a_n=301$: $301=5+(n-1)6\Rightarrow 301=6n-1\Rightarrow n=\dfrac{302}{6}=\dfrac{151}{3}$. Since $n$ is not a positive integer, $301$ is not a term of the list.

NCERT Example 7 — how many two-digit numbers are divisible by 3?

The list is $12,15,18,\dots,99$, an AP with $a=12,\ d=3,\ a_n=99$. So $99=12+(n-1)3\Rightarrow 87=(n-1)3\Rightarrow n-1=29\Rightarrow n=30$. There are 30 such numbers.

NCERT Example 8 — 11th term from the last

AP $10,7,4,\dots,-62$: $a=10,\ d=-3,\ l=-62$. First find total terms: $-62=10+(n-1)(-3)\Rightarrow-72=(n-1)(-3)\Rightarrow n-1=24\Rightarrow n=25$. The 11th term from the last is the $(25-11+1)=$ 15th term: $a_{15}=10+14(-3)=10-42=-32$.

Alternative: reverse the AP — then $a=-62,\ d=+3$, and the 11th term is $a_{11}=-62+10\times3=-62+30=-32$. Same answer.

NCERT Example 9 — does simple interest form an AP?

$\rupee$1000 at $8\%$ simple interest per year: $\text{SI}=\dfrac{P\times R\times T}{100}$. Interests are $80,160,240,\dots$ — differences all $80$, so it is an AP with $a=80,\ d=80$. Interest at the end of 30 years: $a_{30}=80+(30-1)80=80+29\times80=2400$. So $\rupee$2400.

NCERT Example 10 — rows of rose plants

Rows have $23,21,19,\dots,5$ plants: $a=23,\ d=-2,\ a_n=5$. So $5=23+(n-1)(-2)\Rightarrow-18=(n-1)(-2)\Rightarrow n-1=9\Rightarrow n=10$. There are 10 rows.

7. Sum of the first n terms of an AP

Young Gauss summed $1+2+\dots+100$ by writing the sum forwards and backwards and adding:

$$S=1+2+\dots+100,\qquad S=100+99+\dots+1$$$$2S=101\times100\Rightarrow S=\dfrac{100\times101}{2}=5050$$

The same pairing trick on $a,\ a+d,\ a+2d,\dots$ gives the general result. Writing $S$ forwards and backwards and adding term-wise, each of the $n$ pairs equals $2a+(n-1)d$, so $2S=n[2a+(n-1)d]$, hence:

$$\boxed{\,S_n=\dfrac{n}{2}\bigl[2a+(n-1)d\bigr]\,}$$

Since $a+(n-1)d=a_n$ (the last term), we can also write:

$$S_n=\dfrac{n}{2}\bigl[a+a_n\bigr]=\dfrac{n}{2}(a+l)$$

The second form is handy when the first and last terms are known but $d$ is not. The formula links four quantities $S_n,\ a,\ d,\ n$ (or $a_n$) — given any three you can find the fourth. A useful link between sum and term:

$$a_n=S_n-S_{n-1}$$

8. Worked examples on sums

NCERT Example 11 — sum of first 22 terms of $8,3,-2,\dots$

$a=8,\ d=3-8=-5,\ n=22$. So $S_{22}=\dfrac{22}{2}[2(8)+(22-1)(-5)]=11[16+21(-5)]=11(16-105)=11(-89)=-979$.

NCERT Example 12 — find 20th term given $S_{14}=1050,\ a=10$

$1050=\dfrac{14}{2}[2(10)+13d]=7[20+13d]=140+91d\Rightarrow 910=91d\Rightarrow d=10$. Then $a_{20}=10+(20-1)10=200$. The 20th term is 200.

NCERT Example 13 — how many terms of $24,21,18,\dots$ sum to 78?

$a=24,\ d=-3,\ S_n=78$. So $78=\dfrac{n}{2}[48+(n-1)(-3)]=\dfrac{n}{2}[51-3n]$. This gives $3n^2-51n+156=0\Rightarrow n^2-17n+52=0\Rightarrow(n-4)(n-13)=0\Rightarrow n=4$ or $13$. Both are valid: terms 5 through 13 are positive and negative and cancel, so the sum of the first 4 equals the sum of the first 13.

NCERT Example 14 — sum of first $n$ positive integers

(i) First 1000 integers: $S_{1000}=\dfrac{1000}{2}(1+1000)=500\times1001=500500$.

(ii) First $n$ integers: $a=1,\ l=n$, so $S_n=\dfrac{n}{2}(1+n)=\dfrac{n(n+1)}{2}$.

NCERT Example 15 — sum of 24 terms when $a_n=3+2n$

$a_1=3+2=5,\ a_2=7,\ a_3=9,\dots$ so $a=5,\ d=2,\ n=24$. Then $S_{24}=\dfrac{24}{2}[2(5)+(24-1)2]=12[10+46]=12\times56=672$.

NCERT Example 16 — TV-set production

3rd-year production $600$, 7th-year $700$: $a+2d=600$ and $a+6d=700\Rightarrow 4d=100\Rightarrow d=25,\ a=550$.

(i) 1st year $=a=\mathbf{550}$. (ii) 10th year $a_{10}=550+9\times25=\mathbf{775}$. (iii) Total in 7 years $S_7=\dfrac{7}{2}[2(550)+6(25)]=\dfrac{7}{2}[1100+150]=\dfrac{7}{2}\times1250=\mathbf{4375}$.

9. Arithmetic mean

From NCERT's "Note to the reader": if $a,b,c$ are in AP, the middle term is the arithmetic mean of the other two:

$$b=\dfrac{a+c}{2}$$

This is just the AP condition $b-a=c-b$ rearranged. It is a quick way to insert a term between two given numbers, or to check three numbers form an AP.

10. NCERT Exercise 5.1 — fully solved

Q1. Is it an AP?

  • (i) Taxi fare $\rupee15$ first km, $\rupee8$ each extra km: $15,23,31,\dots$ — $d=8$ constant, AP.
  • (ii) Vacuum removes $\tfrac14$ of remaining air: each term is $\tfrac34$ of the previous (multiplied, not added) — not an AP.
  • (iii) Digging cost $\rupee150$ first metre, rising $\rupee50$ each metre: $150,200,250,\dots$ — $d=50$, AP.
  • (iv) $\rupee10000$ at $8\%$ compound interest yearly: amount multiplies by $1.08$ each year — not an AP.

Q2. First four terms.

  • (i) $a=10,d=10$: $10,20,30,40$.
  • (ii) $a=-2,d=0$: $-2,-2,-2,-2$.
  • (iii) $a=4,d=-3$: $4,1,-2,-5$.
  • (iv) $a=-1,d=\tfrac12$: $-1,-\tfrac12,0,\tfrac12$.
  • (v) $a=-1.25,d=-0.25$: $-1.25,-1.50,-1.75,-2.00$.

Q3. First term and common difference.

  • (i) $3,1,-1,-3,\dots$: $a=3,\ d=-2$.
  • (ii) $-5,-1,3,7,\dots$: $a=-5,\ d=4$.
  • (iii) $\tfrac13,\tfrac53,\tfrac93,\tfrac{13}{3},\dots$: $a=\tfrac13,\ d=\tfrac43$.
  • (iv) $0.6,1.7,2.8,3.9,\dots$: $a=0.6,\ d=1.1$.

Q4. Which form an AP? give $d$ and three more terms.

  • (i) $2,4,8,16,\dots$: $4-2=2$ but $8-4=4$ — not an AP.
  • (ii) $2,\tfrac52,3,\tfrac72,\dots$: $d=\tfrac12$, AP; next $4,\tfrac92,5$.
  • (iii) $-1.2,-3.2,-5.2,-7.2,\dots$: $d=-2$, AP; next $-9.2,-11.2,-13.2$.
  • (iv) $-10,-6,-2,2,\dots$: $d=4$, AP; next $6,10,14$.
  • (v) $3,\ 3+\sqrt2,\ 3+2\sqrt2,\ 3+3\sqrt2,\dots$: $d=\sqrt2$, AP; next $3+4\sqrt2,3+5\sqrt2,3+6\sqrt2$.
  • (vi) $0.2,0.22,0.222,0.2222,\dots$: differences $0.02,0.002,\dots$ vary — not an AP.
  • (vii) $0,-4,-8,-12,\dots$: $d=-4$, AP; next $-16,-20,-24$.
  • (viii) $-\tfrac12,-\tfrac12,-\tfrac12,-\tfrac12,\dots$: $d=0$, AP; next $-\tfrac12,-\tfrac12,-\tfrac12$.
  • (ix) $1,3,9,27,\dots$: ratios constant, not differences — not an AP.
  • (x) $a,2a,3a,4a,\dots$: $d=a$, AP; next $5a,6a,7a$.
  • (xi) $a,a^2,a^3,a^4,\dots$: $a^2-a\neq a^3-a^2$ in general — not an AP.
  • (xii) $\sqrt2,\sqrt8,\sqrt{18},\sqrt{32},\dots=\sqrt2,2\sqrt2,3\sqrt2,4\sqrt2,\dots$: $d=\sqrt2$, AP; next $5\sqrt2,6\sqrt2,7\sqrt2$ i.e. $\sqrt{50},\sqrt{72},\sqrt{98}$.
  • (xiii) $\sqrt3,\sqrt6,\sqrt9,\sqrt{12},\dots$: $\sqrt6-\sqrt3\neq\sqrt9-\sqrt6$ — not an AP.
  • (xiv) $1^2,3^2,5^2,7^2,\dots=1,9,25,49,\dots$: differences $8,16,24$ vary — not an AP.
  • (xv) $1^2,5^2,7^2,73,\dots=1,25,49,73,\dots$: differences $24,24,24$ — AP with $d=24$; next $97,121,145$.

11. NCERT Exercise 5.2 — fully solved

Q1. Fill the blanks (using $a_n=a+(n-1)d$).

  • (i) $a=7,d=3,n=8$: $a_8=7+7\times3=28$.
  • (ii) $a=-18,n=10,a_{10}=0$: $0=-18+9d\Rightarrow d=2$.
  • (iii) $d=-3,n=18,a_{18}=-5$: $-5=a+17(-3)\Rightarrow a=46$.
  • (iv) $a=-18.9,d=2.5,a_n=3.6$: $3.6=-18.9+(n-1)2.5\Rightarrow 22.5=(n-1)2.5\Rightarrow n=10$.
  • (v) $a=3.5,d=0,n=105$: $a_{105}=3.5+104\times0=3.5$.

Q2. Choose correct option.

  • (i) 30th term of $10,7,4,\dots$: $a=10,d=-3$; $a_{30}=10+29(-3)=-77$. Answer (C).
  • (ii) 11th term of $-3,-\tfrac12,2,\dots$: $a=-3,d=\tfrac52$; $a_{11}=-3+10\times\tfrac52=22$. Answer (B).

Q3. Missing terms.

  • (i) $2,\_,26$: middle $=\tfrac{2+26}{2}=14$.
  • (ii) $\_,13,\_,3$: here $a_2=13,a_4=3\Rightarrow 2d=-10\Rightarrow d=-5$; so $a=18$ and $a_3=8$. Terms $18,13,8,3$.
  • (iii) $5,\_,\_,9\tfrac12$: $a=5,a_4=\tfrac{19}{2}\Rightarrow 3d=\tfrac{9}{2}\Rightarrow d=\tfrac32$; terms $5,\tfrac{13}{2},8,\tfrac{19}{2}$.
  • (iv) $-4,\_,\_,\_,\_,6$: $a=-4,a_6=6\Rightarrow 5d=10\Rightarrow d=2$; terms $-4,-2,0,2,4,6$.
  • (v) $\_,38,\_,\_,\_,-22$: $a_2=38,a_6=-22\Rightarrow 4d=-60\Rightarrow d=-15$; $a=53$; terms $53,38,23,8,-7,-22$.

Q4. Which term of $3,8,13,18,\dots$ is $78$? $a=3,d=5$; $78=3+(n-1)5\Rightarrow 75=5(n-1)\Rightarrow n=16$.

Q5. Number of terms.

  • (i) $7,13,19,\dots,205$: $a=7,d=6$; $205=7+(n-1)6\Rightarrow n=34$.
  • (ii) $18,15\tfrac12,13,\dots,-47$: $a=18,d=-\tfrac52$; $-47=18+(n-1)(-\tfrac52)\Rightarrow n=27$.

Q6. Is $-150$ a term of $11,8,5,2,\dots$? $a=11,d=-3$; $-150=11+(n-1)(-3)\Rightarrow n=\tfrac{164}{3}+1$, not an integer — no.

Q7. 11th term $=38$, 16th $=73$. $a+10d=38,\ a+15d=73\Rightarrow 5d=35\Rightarrow d=7,\ a=-32$. So $a_{31}=-32+30\times7=178$.

Q8. 50-term AP, 3rd term $12$, last ($a_{50}$) $=106$. $a+2d=12,\ a+49d=106\Rightarrow 47d=94\Rightarrow d=2,\ a=8$. Then $a_{29}=8+28\times2=64$.

Q9. 3rd term $4$, 9th term $-8$. $a+2d=4,\ a+8d=-8\Rightarrow 6d=-12\Rightarrow d=-2,\ a=8$. Zero term: $8+(n-1)(-2)=0\Rightarrow n=5$. The 5th term is zero.

Q10. $a_{17}=a_{10}+7\Rightarrow (a+16d)-(a+9d)=7\Rightarrow 7d=7\Rightarrow d=1$.

Q11. $3,15,27,39,\dots$: $a=3,d=12$. We need $a_n=a_{54}+132$. $a_{54}=3+53\times12=639$, so $a_n=771$; $771=3+(n-1)12\Rightarrow n=65$. The 65th term.

Q12. Two APs with the same $d$; difference of 100th terms is $100$. Since both share $d$, $a_n-b_n=a-b$ is constant, so $a-b=100$. Difference of 1000th terms is also 100.

Q13. Three-digit multiples of $7$: $105,112,\dots,994$; $a=105,d=7$; $994=105+(n-1)7\Rightarrow n=128$.

Q14. Multiples of $4$ between $10$ and $250$: $12,16,\dots,248$; $a=12,d=4$; $248=12+(n-1)4\Rightarrow n=60$.

Q15. $63,65,67,\dots$ has $a=63,d=2$; $3,10,17,\dots$ has $a=3,d=7$. Equal $n$th terms: $63+(n-1)2=3+(n-1)7\Rightarrow 60=5(n-1)\Rightarrow n=13$.

Q16. 3rd term $16$ and $a_7=a_5+12$. $a+2d=16$; $a_7-a_5=2d=12\Rightarrow d=6$, so $a=16-12=4$. AP: $4,10,16,22,\dots$

Q17. 20th term from the last of $3,8,13,\dots,253$. $a=3,d=5,l=253$; from the last $a=253,d=-5$: 20th $=253+19(-5)=158$.

Q18. $a_4+a_8=24$ and $a_6+a_{10}=44$. $(a+3d)+(a+7d)=2a+10d=24$; $(a+5d)+(a+9d)=2a+14d=44$. Subtract: $4d=20\Rightarrow d=5$, then $2a+50=24\Rightarrow a=-13$. First three terms: $-13,-8,-3$.

Q19. Salary $\rupee5000$ in 1995, rising $\rupee200$/year, reaching $\rupee7000$. $7000=5000+(n-1)200\Rightarrow n-1=10\Rightarrow n=11$. Year $=1995+10=$ 2005.

Q20. Weekly saving $\rupee5$, rising $\rupee1.75$, becomes $\rupee20.75$. $20.75=5+(n-1)1.75\Rightarrow 15.75=(n-1)1.75\Rightarrow n-1=9\Rightarrow n=10$.

12. NCERT Exercise 5.3 — fully solved

Q1. Find the sums.

  • (i) $2,7,12,\dots$ to 10 terms: $a=2,d=5$; $S_{10}=\tfrac{10}{2}[4+9\times5]=5\times49=245$.
  • (ii) $-37,-33,-29,\dots$ to 12 terms: $a=-37,d=4$; $S_{12}=\tfrac{12}{2}[-74+11\times4]=6(-30)=-180$.
  • (iii) $0.6,1.7,2.8,\dots$ to 100 terms: $a=0.6,d=1.1$; $S_{100}=\tfrac{100}{2}[1.2+99\times1.1]=50(1.2+108.9)=50\times110.1=5505$.
  • (iv) $\tfrac{1}{15},\tfrac{1}{12},\tfrac{1}{10},\dots$ to 11 terms: $a=\tfrac{1}{15},d=\tfrac{1}{12}-\tfrac{1}{15}=\tfrac{1}{60}$; $S_{11}=\tfrac{11}{2}[\tfrac{2}{15}+10\cdot\tfrac{1}{60}]=\tfrac{11}{2}[\tfrac{8}{60}+\tfrac{10}{60}]=\tfrac{11}{2}\cdot\tfrac{18}{60}=\tfrac{33}{20}$.

Q2. Sums of given series.

  • (i) $7+10\tfrac12+14+\dots+84$: $a=7,d=\tfrac72,l=84$; $84=7+(n-1)\tfrac72\Rightarrow n=23$; $S=\tfrac{23}{2}(7+84)=\tfrac{23}{2}\times91=\tfrac{2093}{2}=1046\tfrac12$.
  • (ii) $34+32+30+\dots+10$: $a=34,d=-2,l=10$; $10=34+(n-1)(-2)\Rightarrow n=13$; $S=\tfrac{13}{2}(34+10)=\tfrac{13}{2}\times44=286$.
  • (iii) $-5+(-8)+(-11)+\dots+(-230)$: $a=-5,d=-3,l=-230$; $-230=-5+(n-1)(-3)\Rightarrow n=76$; $S=\tfrac{76}{2}(-5-230)=38\times(-235)=-8930$.

Q3. In an AP.

  • (i) $a=5,d=3,a_n=50$: $50=5+(n-1)3\Rightarrow n=16$; $S_{16}=\tfrac{16}{2}(5+50)=8\times55=440$.
  • (ii) $a=7,a_{13}=35$: $35=7+12d\Rightarrow d=\tfrac{7}{3}$; $S_{13}=\tfrac{13}{2}(7+35)=\tfrac{13}{2}\times42=273$.
  • (iii) $a_{12}=37,d=3$: $37=a+11\times3\Rightarrow a=4$; $S_{12}=\tfrac{12}{2}(4+37)=6\times41=246$.
  • (iv) $a_3=15,S_{10}=125$: $a+2d=15$; $S_{10}=5(2a+9d)=125\Rightarrow 2a+9d=25$. Solve: from $a=15-2d$, $30-4d+9d=25\Rightarrow 5d=-5\Rightarrow d=-1,\ a=17$; $a_{10}=17+9(-1)=8$.
  • (v) $d=5,S_9=75$: $S_9=\tfrac{9}{2}(2a+8\times5)=75\Rightarrow 2a+40=\tfrac{150}{9}=\tfrac{50}{3}\Rightarrow a=\tfrac{50/3-40}{2}=-\tfrac{35}{3}$; $a_9=a+8\times5=-\tfrac{35}{3}+40=\tfrac{85}{3}$.
  • (vi) $a=2,d=8,S_n=90$: $90=\tfrac{n}{2}[4+(n-1)8]\Rightarrow 180=4n+8n^2-8n\Rightarrow 8n^2-4n-180=0\Rightarrow 2n^2-n-45=0\Rightarrow(2n+9)(n-5)=0\Rightarrow n=5$; $a_5=2+4\times8=34$.
  • (vii) $a=8,a_n=62,S_n=210$: $210=\tfrac{n}{2}(8+62)=35n\Rightarrow n=6$; $62=8+5d\Rightarrow d=\tfrac{54}{5}=10.8$.
  • (viii) $a_n=4,d=2,S_n=-14$: $4=a+(n-1)2\Rightarrow a=6-2n$; $-14=\tfrac{n}{2}(a+4)=\tfrac{n}{2}(10-2n)\Rightarrow -14=5n-n^2\Rightarrow n^2-5n-14=0\Rightarrow(n-7)(n+2)=0\Rightarrow n=7,\ a=6-14=-8$.
  • (ix) $a=3,n=8,S=192$: $192=\tfrac{8}{2}(6+7d)=4(6+7d)\Rightarrow 48=6+7d\Rightarrow d=6$.
  • (x) $l=28,S=144,n=9$: $144=\tfrac{9}{2}(a+28)\Rightarrow 32=a+28\Rightarrow a=4$.

Q4. $9,17,25,\dots$ sum to $636$: $a=9,d=8$; $636=\tfrac{n}{2}[18+(n-1)8]\Rightarrow 1272=18n+8n^2-8n\Rightarrow 8n^2+10n-1272=0\Rightarrow 4n^2+5n-636=0\Rightarrow(n-12)(4n+53)=0\Rightarrow n=12$.

Q5. $a=5,l=45,S=400$: $400=\tfrac{n}{2}(5+45)=25n\Rightarrow n=16$; $45=5+15d\Rightarrow d=\tfrac{40}{15}=\tfrac{8}{3}$.

Q6. $a=17,l=350,d=9$: $350=17+(n-1)9\Rightarrow n=38$; $S=\tfrac{38}{2}(17+350)=19\times367=6973$.

Q7. $d=7,a_{22}=149$: $149=a+21\times7\Rightarrow a=2$; $S_{22}=\tfrac{22}{2}(2+149)=11\times151=1661$.

Q8. $a_2=14,a_3=18\Rightarrow d=4,\ a=10$; $S_{51}=\tfrac{51}{2}[20+50\times4]=\tfrac{51}{2}\times220=5610$.

Q9. $S_7=49,\ S_{17}=289$. $\tfrac72(2a+6d)=49\Rightarrow a+3d=7$; $\tfrac{17}{2}(2a+16d)=289\Rightarrow a+8d=17$. Subtract: $5d=10\Rightarrow d=2,\ a=1$. $S_n=\tfrac{n}{2}[2+(n-1)2]=n^2$.

Q10. Show these are APs and find $S_{15}$.

  • (i) $a_n=3+4n$: $a_{n+1}-a_n=4$ constant — AP, $a_1=7,d=4$; $S_{15}=\tfrac{15}{2}[14+14\times4]=\tfrac{15}{2}\times70=525$.
  • (ii) $a_n=9-5n$: difference $-5$ — AP, $a_1=4,d=-5$; $S_{15}=\tfrac{15}{2}[8+14(-5)]=\tfrac{15}{2}(-62)=-465$.

Q11. $S_n=4n-n^2$. $S_1=3$ so $a_1=3$; $S_2=4$ so $a_2=S_2-S_1=1$, $a_3=S_3-S_2=3-4=-1$, $a_{10}=S_{10}-S_9=-60-(-45)=-15$. General $a_n=S_n-S_{n-1}=(4n-n^2)-(4(n-1)-(n-1)^2)=5-2n$.

Q12. First 40 positive integers divisible by 6: $6,12,\dots$; $a=6,d=6,n=40$; $S_{40}=\tfrac{40}{2}[12+39\times6]=20(12+234)=20\times246=4920$.

Q13. First 15 multiples of 8: $a=8,d=8,n=15$; $S_{15}=\tfrac{15}{2}[16+14\times8]=\tfrac{15}{2}\times128=960$.

Q14. Odd numbers between 0 and 50: $1,3,\dots,49$; $a=1,d=2,n=25$; $S_{25}=\tfrac{25}{2}(1+49)=25\times25=625$.

Q15. Penalty $200,250,300,\dots$ for 30 days: $a=200,d=50,n=30$; $S_{30}=\tfrac{30}{2}[400+29\times50]=15(400+1450)=15\times1850=27750$. Penalty $\rupee$27750.

Q16. Seven prizes summing to $\rupee700$, each $\rupee20$ less than the previous. $S_7=700,n=7,d=-20$: $700=\tfrac72(2a+6(-20))\Rightarrow 200=2a-120\Rightarrow a=160$. Prizes: $\rupee160,140,120,100,80,60,40$.

Q17. Each section plants trees equal to its class number, 3 sections each, Classes I–XII. Trees $=3(1+2+\dots+12)=3\times\tfrac{12\times13}{2}=3\times78=234$.

Q18. Spiral of 13 semicircles, radii $0.5,1.0,1.5,\dots$ cm. Length $=\pi(r_1+r_2+\dots+r_{13})$, radii in AP $a=0.5,d=0.5,n=13$: sum $=\tfrac{13}{2}[1+12\times0.5]=\tfrac{13}{2}\times7=45.5$. Length $=\tfrac{22}{7}\times45.5=143$ cm.

Q19. Logs $20,19,18,\dots$ totalling 200. $a=20,d=-1,S_n=200$: $200=\tfrac{n}{2}[40-(n-1)]\Rightarrow 400=41n-n^2\Rightarrow n^2-41n+400=0\Rightarrow(n-16)(n-25)=0$. $n=25$ gives $a_{25}=20-24=-4$ (impossible), so $n=16$; top row $a_{16}=20-15=5$ logs. 16 rows, 5 logs on top.

Q20. Potato race: distances $2\times5,\ 2\times8,\ 2\times11,\dots$ for 10 potatoes $=10,16,22,\dots$; $a=10,d=6,n=10$; $S_{10}=\tfrac{10}{2}[20+9\times6]=5\times74=370$ m.

13. NCERT Exercise 5.4 (Optional) — fully solved

Q1. $121,117,113,\dots$: $a=121,d=-4$. First negative term needs $a_n<0$: $121+(n-1)(-4)<0\Rightarrow 125<4n\Rightarrow n>31.25\Rightarrow n=32$. The 32nd term is the first negative one.

Q2. $a_3+a_7=6$ and $a_3\cdot a_7=8$. So $(a+2d)+(a+6d)=2a+8d=6\Rightarrow a+4d=3$. The two terms multiply to $8$ and add to $6$, so they are $2$ and $4$. If $a+2d=2,a+6d=4\Rightarrow d=\tfrac12,a=1$; then $S_{16}=\tfrac{16}{2}[2+15\times\tfrac12]=8\times9.5=76$. (Other case $d=-\tfrac12,a=5$ gives $S_{16}=\tfrac{16}{2}[10-7.5]=20$.)

Q3. Ladder rungs 25 cm apart, length from 45 cm (bottom) to 25 cm (top), rungs span $2\tfrac12$ m $=250$ cm. Number of rungs $=\tfrac{250}{25}+1=11$. Lengths form an AP $a=45,l=25,n=11$. Total wood $=S_{11}=\tfrac{11}{2}(45+25)=\tfrac{11}{2}\times70=385$ cm.

Q4. Houses $1$ to $49$; find $x$ with sum before $x$ = sum after $x$. So $S_{x-1}=S_{49}-S_x$, i.e. $\tfrac{(x-1)x}{2}=\tfrac{49\times50}{2}-\tfrac{x(x+1)}{2}$. This gives $x^2=\tfrac{49\times50}{2}=1225\Rightarrow x=35$. The house is numbered 35.

Q5. Terrace of 15 steps, each $50$ m long, rise $\tfrac14$ m, tread $\tfrac12$ m. Volume of step $k$ (counting from top) $=\tfrac14\times\tfrac12\times50\times k=\tfrac{50}{8}k$ m$^3$. Total $=\tfrac{50}{8}(1+2+\dots+15)=\tfrac{50}{8}\times\tfrac{15\times16}{2}=\tfrac{50}{8}\times120=750$ m$^3$.

14. Common mistakes to avoid

  • Computing $d$ as $a_k-a_{k+1}$ instead of $a_{k+1}-a_k$ — keep "later minus earlier".
  • Using $a+nd$ for the $n$th term — it is $a+(n-1)d$, with $(n-1)$.
  • Forgetting that a list with a constant ratio (like $1,3,9,27$) is not an AP.
  • Accepting a non-integer $n$ as an answer — the number of terms must be a positive whole number (so "is X a term?" is answered "no").
  • For sums, mixing up $S_n=\tfrac{n}{2}[2a+(n-1)d]$ with $\tfrac{n}{2}(a+l)$ — both are correct, just pick the one matching the given data.
  • Rejecting a valid second root when an AP turns from positive to negative (Example 13 has two answers).
  • In "term from the last", the $k$th term from the end is the $(n-k+1)$th from the start.

15. Quick revision checklist

  • AP: constant difference $d=a_{k+1}-a_k$; general form $a,a+d,a+2d,\dots$
  • $n$th term: $a_n=a+(n-1)d$.
  • Sum: $S_n=\dfrac{n}{2}[2a+(n-1)d]=\dfrac{n}{2}(a+a_n)$.
  • $a_n=S_n-S_{n-1}$; arithmetic mean $b=\dfrac{a+c}{2}$.
  • Two unknowns? Make two equations from two given terms or sums, then solve.
  • $d>0$ increasing, $d<0$ decreasing, $d=0$ constant.
Practice MCQs
1. The common difference of the AP $\tfrac13,\tfrac{1-3b}{3},\tfrac{1-6b}{3},\dots$ is:
  1. $\tfrac13$
  2. $-b$
  3. $b$
  4. $-\tfrac{b}{3}$
Answer: (B) $d=\tfrac{1-3b}{3}-\tfrac13=\tfrac{-3b}{3}=-b.$
2. The 10th term of the AP $2,7,12,\dots$ is:
  1. $45$
  2. $47$
  3. $52$
  4. $42$
Answer: (B) $a_{10}=2+9\times5=47.$
3. Which term of the AP $21,18,15,\dots$ is $0$?
  1. 6th
  2. 7th
  3. 8th
  4. 9th
Answer: (C) $21+(n-1)(-3)=0\Rightarrow n=8.$
4. If the $n$th term of an AP is $3+2n$, its common difference is:
  1. $3$
  2. $2$
  3. $5$
  4. $n$
Answer: (B) consecutive terms differ by the coefficient of $n$, i.e. $2.$
5. The sum of the first $n$ positive integers is:
  1. $\tfrac{n(n+1)}{2}$
  2. $n^2$
  3. $\tfrac{n(n-1)}{2}$
  4. $n(n+1)$
Answer: (A) $S_n=\tfrac{n}{2}(1+n)=\tfrac{n(n+1)}{2}.$
6. The sum of the first 22 terms of $8,3,-2,\dots$ is:
  1. $-979$
  2. $979$
  3. $-880$
  4. $-1000$
Answer: (A) $S_{22}=11(16-105)=-979.$
7. How many two-digit numbers are divisible by 3?
  1. $28$
  2. $29$
  3. $30$
  4. $33$
Answer: (C) $12,15,\dots,99$ gives $n=30.$
8. If $a,b,c$ are in AP, then $b$ equals:
  1. $ac$
  2. $\tfrac{a+c}{2}$
  3. $\sqrt{ac}$
  4. $a+c$
Answer: (B) the middle term is the arithmetic mean, $b=\tfrac{a+c}{2}.$
9. The 11th term from the last of $10,7,4,\dots,-62$ is:
  1. $-25$
  2. $-32$
  3. $-29$
  4. $-35$
Answer: (B) reversed AP $a=-62,d=3$; $a_{11}=-62+30=-32.$
10. The first negative term of the AP $121,117,113,\dots$ is the:
  1. 30th
  2. 31st
  3. 32nd
  4. 33rd
Answer: (C) $121-4(n-1)<0\Rightarrow n>31.25\Rightarrow n=32.$
11. If $S_n=4n-n^2$, then $a_2$ equals:
  1. $3$
  2. $1$
  3. $-1$
  4. $4$
Answer: (B) $a_2=S_2-S_1=4-3=1.$
12. Which list is an AP?
  1. $2,4,8,16,\dots$
  2. $1,3,9,27,\dots$
  3. $\sqrt2,\sqrt8,\sqrt{18},\dots$
  4. $0.2,0.22,0.222,\dots$
Answer: (C) $=\sqrt2,2\sqrt2,3\sqrt2,\dots$ with $d=\sqrt2.$
Assertion–Reason
A: The list $1,3,9,27,\dots$ is not an AP.   R: In an AP the difference between consecutive terms is constant.
Answer: Both A and R are true, and R correctly explains A — here the ratio is constant, not the difference, so it fails the AP test.
A: The 8th term of $21,18,15,\dots$ is $0$.   R: The $n$th term of an AP is $a_n=a+nd$.
Answer: A is true but R is false — the correct formula is $a_n=a+(n-1)d$. (Using it, $a_8=21+7(-3)=0$.)
Previous-year questions
Q1. How many terms of the AP $9,17,25,\dots$ must be taken to give a sum of $636$? (CBSE, 3 marks)
Answer: $a=9,d=8$; $636=\tfrac{n}{2}[18+8(n-1)]\Rightarrow 4n^2+5n-636=0\Rightarrow(n-12)(4n+53)=0\Rightarrow n=12$.
Q2. 200 logs are stacked with 20 in the bottom row, 19 in the next, and so on. In how many rows are they placed and how many are in the top row? (CBSE, 3 marks)
Answer: $a=20,d=-1$; $200=\tfrac{n}{2}[41-n]\Rightarrow n^2-41n+400=0\Rightarrow n=16$ (reject $25$). Top row $a_{16}=20-15=5$ logs. 16 rows, 5 on top.
Q3. The 17th term of an AP exceeds its 10th term by 7. Find the common difference. (CBSE, 2 marks)
Answer: $(a+16d)-(a+9d)=7\Rightarrow 7d=7\Rightarrow d=1.$
Q4. If the sum of the first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of its first $n$ terms. (CBSE, 3 marks)
Answer: $a+3d=7$ and $a+8d=17\Rightarrow d=2,a=1$; $S_n=\tfrac{n}{2}[2+(n-1)2]=n^2$.
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More Class 10 Mathematics chapters
Real Numbers · Polynomials · Pair of Linear Equations in Two Variables · Quadratic Equations · Triangles · Coordinate Geometry · Introduction to Trigonometry · Some Applications of Trigonometry