NCERT Solutions for Class 9 Maths Chapter 4 Exploring Algebraic Identities

NCERT Solutions for Class 9 Maths Chapter 4 help students strengthen their understanding of Algebraic Identities concepts and build the confidence needed to score well in exams. These solutions align with the latest NCERT syllabus and the CBSE curriculum, ensuring students study from content that stays fully exam-relevant.

Access Detailed answers for every Important question in the chapter 4, and they download can the complete solution set as a free PDF for offline study. Subject Experts at ALLEN designs each solution to help students follow the correct method, avoid common mistakes, and revise efficiently before exams.

1.0Download NCERT Class 9 Maths Chapter 4 Exploring Algebraic Identities

Class 9 Maths Chapter 4: Exploring Algebraic Identities introduces students to algebraic identities and their applications in simplifying algebraic expressions and solving mathematical problems. Understanding these identities helps build a strong foundation for higher algebra. Download NCERT Solutions for Class 9 Maths Chapter 4 PDF to access step-by-step answers based on the latest NCERT syllabus 2026-27 and prepare confidently for your exams

NCERT Solutions Class 9 Maths Chapter 4 Algebraic Identities

2.0Learning Outcomes - NCERT Class 9 Maths Chapter 4 Exploring Algebraic Identities

Understand the concept of algebraic identities and distinguish them from algebraic equations.
Visualize algebraic identities using geometric models and algebra tiles.
Apply standard algebraic identities to expand algebraic expressions.
Factorize algebraic expressions using suitable identities.
Discover and use new algebraic identities to solve mathematical problems.
Simplify rational algebraic expressions using factorization techniques.
Solve application-based and NCERT textbook questions confidently using algebraic identities.

3.0Detailed Class 9 Maths Chapter 4 Exploring Algebraic Identities Solutions

Exercise : 4.1

1.Find the first five terms of the sequence in which the $n^{th}$ term is given by (i) $t_{n} = 3 n - 4$ (ii) $t_{n} = 2 - 5 n$ (iii) $t_{n} = n^{2} - 2 n + 3$ for $n \geq 1$

Sol. (i) Since $t_{n} = 3 n - 4$

$For n = 1; t_{1} = 3 (1) - 4 = 3 - 4 = - 1 n = 2; t_{2} = 3 (2) - 4 = 6 - 4 = 2 n = 3; t_{3} = 3 (3) - 4 = 9 - 4 = 5 n = 4; t_{4} = 3 (4) - 4 = 12 - 4 = 8 n = 5; t_{5} = 3 (5) - 4 = 15 - 4 = 11$

(ii) $t_{n} = 2 - 5 n$

2.Determine whether 97 and 172 are terms of the sequence $t_{n} = 5 n - 3$ for $n \geq 1$ .

Sol. If 97 and 172 are terms of the sequence $t_{n} = 5 n - 3$ then ' $n$ ' should be natural number or positive integer for these given values.

Now, $5 n - 3 = 97$ 5n = 100 n = 20 Yes 97 is term of given sequence. again $5 n - 3 = 172$ 5n $= 175$ $n = \frac{175}{5}$ n = 35 Yes 172 is also a term of given sequence. 3. Which term of the sequence $t_{n} = 5 n - 3$ for $n \geq 1$ is 607 ?

Sol. $∵ t_{n} = 5 n - 3$

So, $5 n - 3 = 607$ If 607 is term of given sequence $5 n = 610$ $n = \frac{610}{5}$ $n = 122$ thus, given term will occur at $122^{th}$ place. 4. A sequence is given by the recursive rule $t_{1} = - 5, t_{n + 1} = t_{n} + 3$ for $n \geq 1$ . Find the first five terms of the sequence. Is 52 a term of this sequence? If so, which term is it? (iii) $t_{n} = n^{2} = 2 n + 3$ for $n \geq 1$

Sol. Since $t_{1} = - 5 t_{n + 1} = t_{n} + 3$ for $n \geq 1$ if $n = 1$ then $t_{1 + 1} = t_{1} + 3$ $t_{2} = - 5 + 3$ $t_{2} = - 2$ If $n = 2$ then $t_{3} = t_{2} + 3$ $t_{3} = - 2 + 3 = 1$ $t_{3} = 1$ if $n = 3$ then $t_{4} = t_{3} + 3$ $t_{4} = 1 + 3 = 4$ $t_{4} = 4$ if $n = 4$ then $t_{5} = t_{4} + 3$ $t_{5} = 4 + 3 = 7$ $t_{5} = 7$ So first five terms are $= - 5, - 2, 1, 4, 7$ This is an arithmetic sequence with first term $a = - 5$ common difference $d = 3$

Since Formula $t_{n} = a + (n - 1) d$ $t_{n} = - 5 + (n - 1) 3$ $t_{n} = - 5 + 3 n - 3$ $n \geq 1$ Now checking whether 52 is a term Let $t = 52$ $3 n - 8 = 52$ $3 n = 60$ n = 20 It is a natural number. Therefore, 52 is the $2 0^{th}$ term. 5. Let $T_{1} = 1, T_{2} = 2, T_{3} = 4$ and $T_{n} = T_{n - 1} + T_{n - 2} + T_{n - 3}$ (for $n \geq 1$ ). Find $T_{4}, T_{5}, T_{6}, T_{7}$ , and $T_{8}$ . Sol. $T_{1} = 1$ $T_{2} = 2$ $T_{3} = 4$ $T_{n} = T_{n - 1} + T_{n - 2} + T_{n - 3} ($ for $n \geq 4)$ Now $T_{4} = T_{3} + T_{2} + T_{1}$ (Put $n = 4$ in above sequence format) $T_{4} = 4 + 2 + 1$ $T_{4} = 7$ $T_{5} = T_{4} + T_{3} + T_{2}$ (use $n = 5$ ) $T_{5} = 7 + 4 + 2$ $T_{5} = 13$ $T_{6} = T_{5} + T_{4} + T_{3}$ (use $n = 6$ ) $T_{6} = 13 + 7 + 4$ $T_{6} = 24$ $T_{7} = T_{6} + T_{5} + T_{4}$ $T_{7} = 24 + 13 + 7$ $T_{7} = 44$ $T_{8} = T_{7} + T_{6} + T_{5}$ $T_{8} = 44 + 24 + 13$ $T_{8} = 81$

Exercise: 4.2

1.Find the $1 0^{th}$ and $26^{th}$ terms of the AP: 3, 8, 13, 18, ...... Sol. Given A.P. $= 3, 8, 13, 18, \dots\dots$ I $^{st}$ term a $= 3$ Common difference $d = 8 - 3 = 5$

2.Which term of the AP : $21, 18, 15$ , $____$ is -81? Also, is 0 a term of this AP? Give reasons for your answer. Sol. Given A.P. $= 21, 18, 15$ $a = 21; d = 18 - 21 = - 3$ $t_{n} = a + (n - 1) d$ $t_{n} = 21 - 3 n + 3$ $- 81 = 24 - 3 n$ $- 81 - 24 = - 3 n$ $n = \frac{- 105}{- 3} = 35; n = 35$ Now $t_{n} = 0$ $24 - 3 n = 0$ $3 n = 24$ $n = 8$ so, 0 is $8^{th}$ term

3.Find the $n^{th}$ term of the AP: $11, 8, 5, 2$ $____$ Write the recursive rule for this AP. Sol. Given A.P. $= 11, 8, 5, 2$ $a = 11, d = - 3$ $t_{n} = 11 + (n - 1) (- 3)$ $t_{n} = 11 - 3 n + 3$ $t_{n} = 14 - 3 n$ Recursive Rule $t_{1} = 11$ $t_{n} = t_{n - 1} - 3$ for $n \geq 2$

4.An AP consists of 50 terms in which the $3^{rd}$ term is 12 and the last term is 106. Find the $2 9^{th}$ term. Sol. Let first term $= a$ & common difference $= d$ $t_{3} = 12$

$a + 2 d = 12$

AP has 50 terms and last term = 106 $5 0^{th}$ term $= 106$

$a + 49 d = 106$

Equation (2) - (1) $(a + 49 d) - (a + 2 d) = 106 - 12$ 47d = 94 $d = 2$ from equation (1) $a + 2 (2) = 12$ $a = 8$ now find $29^{th}$ term $t_{29} = a + 28 d = 8 + (28 \times 2) = 8 + 56$ $a_{29} = 64$ 5. How many 2-digit numbers are divisible by 3 ? What is the sum of all these 2-digit numbers?

Sol. AP will be $12, 15, 18$ $____$ 99 $a = 12, d = 15 - 12 = 3$ , last term = 99 Now $n =$ ? $t_{n} = a + (n - 1) d$ $99 = 12 + (n - 1) 3$ $= 11, 8, 5, 2$ $(n - 1) = \frac{87}{3}$ $n - 1 = 29$ $n = 30$ So, $S_{30} = \frac{30}{2} [a + a_{n}]$ $S_{30} = 15 (12 + 99)$ $S_{30} = 15 \times 111$ $S_{30} = 1665$ So, such total numbers are 30 and their sum is 1665. 6. Harish started work at an annual salary of ₹5,00,000 and received an increment of ₹20,000 each year. After how many years did his income reach ₹7,00,000?

Sol. Initial salary $=$ ₹ 5,00,000 Annual Increment $= ₹20, 000$ A.P. $= 500000, 520000, 540000$

Using formula $t_{n} = a + (n - 1) d$ $700000 = 500000 + (n - 1) 20000$ $(n - 1) 20000 = 200000$ n - 1 = 10 n = 11 So, Harish’s income will be ₹ 7,00000 Per annum in $11^{th}$ year. So, It took 10 year or 10 yearly increments. 7. A child arranges marbles in rows so that the first row has 1 marble, the second has 2 marbles, the third has 3, and so on up to 25 rows. How many marbles does the child use in all?

Sol. Number of marbles $= 1 + 2 + 3 + \dots\dots + 25$ $S_{n} = \frac{n ( n + 1 )}{2}$ $S_{25} = \frac{25 ( 25 + 1 )}{2}$ $S_{25} = \frac{25 \times 26}{2} = 325$ $∴$ total marbles used by child $= 325$

Exercise : 4.3

1.Find the $1 2^{th}$ term of a GP with common ratio 2 , whose $8^{th}$ term is 192.

Sol. common ratio $r = 2$ $t_{8} = a_{8} (8^{th}$ term $) = 192$ So $t_{8} = a \cdot r^{7} = 192$ $a (2)^{7} = 192$ $n - 1 = 29$ Now $t_{12} = ar^{11}$ $t_{12} = \frac{3}{2} \times (2)^{11}$ $t_{12} = 3 \times 2^{10} = 3072$

2. Find the $1 0^{th}$ and $n^{th}$ terms of the GP: $5, 25, 125, \dots .$ . Sol. G.P. $= 5, 25, 125$ $a = 5$ $r = \frac{a _{2}}{a _{1}} = \frac{25}{5} = 5$ $t^{n} = 5 \times 5^{n - 1} = 5^{n}$ Hence nth term $= 5^{n}$ 10 term $_{10} = 5^{10} = 9765625$

3. A sequence is given by the recursive rule $t_{1} = 2, t_{n + 1} = 3 t_{n} - 2$ for $n \geq 1$ . Which term of the sequence is 730 ?

Sol. $t_{1} = 2$ $r = 2$ If $t = 1$ $t_{2} = 3 t_{1} - 2$ $t_{2} = 3 (2) - 2 = 4$ Same as $t_{3} = 3 t_{2} - 2$ $t_{2} = 3 (4) - 2 = 10$ $t_{4} = 3 t_{3} - 2 = 3 (10) - 2 = 28$ $t_{5} = 3 t_{4} - 2 = 3 (28) - 2 = 82$ $n^{th}$ $t_{7} = 3 t_{6} - 2 = 3 (244) - 2 = 730$

4. Which term of the GP: $2, 6, 18, \dots\dots$ is 4374? Write the explicit formula as well as the recursive formula for the $t^{n} = 5 \times 5^{n - 1} = 5^{n}$ term.

Sol. Given G.P. $= 2, 6, 18 \dots$ $a = 2, r = \frac{a _{2}}{a _{1}} = \frac{6}{2} = 3$ Use formula: $t_{n} = ar^{n - 1}$ $t_{n} = 2 \times 3^{n - 1}$ $t_{n} = 4374$ $2 \times 3^{n - 1} = 4374$ $3^{n - 1} = 2187$ $3^{n - 1} = 37$ $n = 8$ Recursive formula : $t_{1} = 2, t_{n} = 3 t_{n - 1}$ for $n \geq 2$

5. A ball is dropped from a height of 80 metres. After hitting the ground, it bounces back to 60% of the height from which it fell. It continues bouncing in this way-each time rising to $60%$ of the previous height. (i) What height does the ball reach after the $5^{th}$ bounce? (ii) What is the total vertical distance the ball has travelled by the time it hits the ground for the $6^{th}$ time?

Sol. (i) Initial height $= 80 m$ Bounce Ratio $= 60% = 0.6$ Height office $1^{st}$ bounce $= 80 \times 0.6 = 48 m$ Height after $2^{nd}$ bounce $= 48 \times 0.6 = 28.8 m$ . Height after $3^{rd}$ bounce $= 28.8 \times 0.6 = 17.28 m$ Height after $4^{th}$ bounce $= 17.28 \times 0.6 = 10.638 m$ Height after $5^{th}$ bounce $= 10.638 \times 0.6 = 6.2208 m$ (ii) The ball first balls 80 m to hit the ground for the $1^{st}$ time. Then it rises and falls after each bounce.

Height after bounces: $1^{st}$ bounce $= 48 m$ $2^{nd}$ bounce $= 28.8 m$ $3^{rd}$ bounce $= 17.28 m$ $4^{th}$ bounce $= 10.638 m$ $5^{th}$ bounce $= 6.2208 m$ To hit the ground for the $6^{th}$ time, it travels the total distance $= 80 + 2 (48 + 28.8 + 17.28 + 10.638 +$ 6.2208) $t_{n} = 2 \times 3^{n - 1}$ $= 80 + 221.3376$ = 301.3376 Therefore, the total vertical distance travelled in 301.3376 m .

6. Which term of the sequence $2, 22, 4 \dots\dots$ is 128 ?

Sol. $2, 22, 4 \dots\dots\dots$ G.P. $a = 2$ $r = \frac{a _{2}}{a _{1}} = \frac{2 2}{2} = 2$ So, $t_{n} = 2 (2)^{n - 1}$ Use $t_{n} = a \cdot r^{n - 1}$ $t_{n} = 128$ $2 (2)^{n - 1} = 128$ $(2)^{n - 1} = 64$ $(2)^{\frac{n - 1}{2}} = 2^{6}$ $n - 1 = 12$ n = 13 So, 128 is $13^{th}$ term.

7. Figure Stages 0 to 3 of the Sierpinski square carpet. Stage 0 of this fractal is a square sheet of paper. To construct Stage 1, each side of the square is trisected and the points of trisection of opposite sides are joined to obtain nine smaller squares. The centre square is then removed and the 8 smaller squares are retained, leaving a square hole in the centre. The same process is repeated on the eight smaller shaded squares to obtain Stage 2 and so on. (i) How many red squares are there in Stages 0 to 3? (ii) Can you predict the number of red squares in Stages 4 and 5? (iii) Can you find a rule for the number of red squares at the $n^{th}$ stage? Write the explicit formula as well as the recursive formula for the number of red squares at any stage. (iv) Suppose the area of the square in Stage 0 is 1 square unit. What is the area of the red region in Stages 1, 2 and 3 ? What will be the area of the red region in Stages 4 and 5? Find the explicit as well as the recursive formula for the area of the red region

Stage 0

at the $n^{th}$ stage. What happens to this area as n , the number of stages, goes on increasing?

Stage 1

Stage 2

Stage 3

Sol. (i) Stage $0 \to 1$ red square Stage $1 \to 8$ red squares Stage $2 \to 8^{2} = 64$ red squares Stage $3 \to 8^{3} = 512$ red squares Therefore stages 0 to 3 have $1, 8, 64$ and 512 red squares respectively. (ii) The number of red squares is multiplied by 8 at each stage. Stage $4 = 8^{4} = 4096$ Stage $5 = 8^{5} = 32768$ Therefore Stage 4 has 4096 red squares. Stage 5 has 32768 red squares. (iii) At each stage, every red square is replaced by 8 smaller red squares. So the number of red squares forms the sequence. 1, 8, 64, 512...... Explicit the formula $t_{n} = 8^{n}$ . Here stage 0 corresponds to $n = 0$ Recursive formula $t_{0} = 1$ $t_{n} = 8 t_{n - 1}$ , for $n \geq 1$ (iv) At each stage the square is divided into 9 equal parts and the centre part is removed.

So, the remaining red area becomes $\frac{8}{9}$ of the previous area

Stage 0 area $= 1$ square unit Stage 1 area $= \frac{8}{9}$ Stage 2 area $= (\frac{8}{9})^{2} = \frac{64}{81}$ Stage 3 area $= (\frac{8}{9})^{3} = \frac{512}{729}$ Stage 4 area $= (\frac{8}{9})^{4} = \frac{4096}{6561}$ Stage 5 area $= (\frac{8}{9})^{5} = \frac{32768}{59049}$ Explicit formula $A_{n} = (\frac{8}{9})^{n}$ Here stage 0 corresponding to $n = 0$ Recursive formula $A_{0} = 1$ $A_{n} = (\frac{8}{9}) A_{n - 1}$ As ' $n$ ' increases, the area keeps decreasing and gets closer to 0 , but it never becomes exactly 0 .

End of Chapter Exercise

1. Find the $3 1^{st}$ term of an AP whose $1 1^{th}$ term is 38 and $16^{th}$ term is 73 .

Sol. First term $= a$ , common difference $= d$

$1 1^{th} term = a + 10 d = 38 1 6^{th} term = a + 15 d = 73$

Subtracting Equation. (1) from (2)

$(a + 15 d) - (a + 10 d) = 73 - 38 5 d = 35 d = 7 a + 10 (7) = 38 a = - 32$

Now $a_{31} = t_{31} = a + 30 d$

2. Determine the AP whose third term is 16 and whose $7^{th}$ term exceeds the $5^{th}$ term by 12 .

Sol. First term $= a$ , common difference $= d$

$3^{rd} term = a + 2 d = 16$

According to question $7^{th}$ term $- 5^{th}$ term $= 12$ $a + 6 d - (a + 4 d) = 12$ $2 d = 12$ $d = 6$ Now a $+ 2 (6) = 16$ $a = 16 - 12$ $a = 4$ So A.P. 4, 10, 16, 22, 28

3. How many three-digit numbers are divisible by 7 ?

Sol. Smallest three-digit number divisible by $7 = 105$

Largest three-digit number divisible by $7 = 994$

Required A.P. 105, 112, 119,........ 994 Let $t_{n} = 994$ $105 + (n - 1) 7 = 994$ $(n - 1) = \frac{994 - 105}{7}$ n - $1 = 127$ n = 128 Therefore 128, three-digit numbers are divisible by 7 .

4. How many multiples of 4 lie between 10 and 250 ?

Sol. The smallest and largest required multiples 12, 248 required A.P.

12, 16, 20, ...... 248 $3 1^{st}$ $d = 4$ last term $= 248$ $16^{th}$ $12 + (n - 1) 4 = 248$ $n - 1 = \frac{236}{4}$ $n - 1 = 59$ n = 60 Therefore 60 multiples will lie between 10 and 250.

5. Find a GP for which the sum of the first two terms is -4 and the fifth term is 4 times the third term.

Sol. required G.P. $= a, ar, ar^{2}, ar^{3}, ar^{4}, \dots\dots$ Given sum of first two terms $= - 4$ $a + ar = - 4$ $a (1 + r) = - 4$

Also $5^{th}$ term is 4 times the $a_{31} = t_{31} = a + 30 d$ term. $5^{th}$ term $=$ ar $^{4}$ $3^{rd}$ term $=$ ar $^{2}$ $= a$ $r^{2} = 4$ So $r = \pm 2$ Case 1: when $r = 2$ From Eq (1) $a (1 + 2) = - 4$ $3 a = - 4$ $a = - \frac{4}{3}$ Hence G.P. $= \frac{- 4}{3}, \frac{- 8}{3}, \frac{- 16}{3}, \frac{- 32}{3}, \frac{- 64}{3} \dots\dots .$ . Case 2: when $r = 2$ From Eq. (1) $a (1 - 2) = - 4$ $a = 4$ Hence G.P. $= 4, - 8, 16, - 32, 64 \dots\dots$ Therefore, possible GP are $\frac{- 4}{3}, \frac{- 8}{3}, \frac{- 16}{3}, \frac{- 32}{3}$ and $4, - 8, 16, - 32, \dots\dots$

6. Find all possible ways of expressing 100 as the sum of consecutive natural numbers.

Sol. We need consecutive natural numbers whose sum is 100 .

Possible ways 1 term: 100 So, 100 = 100 5 terms Let the numbers be $x - 2, x - 1, x, x + 1, x + 2$ Sum $= 5 x$ $5 x = 100$ $x = 20$ So, $100 = 18 + 19 + 20 + 21 + 22$ 8 terms Let the numbers be $x, x + 1, x + 2, \dots\dots x + 7$ Sum $= 8 x + 28$ $8 x + 28 = 100$ $8 x = 72$ $x = 9$ Thus, $100 = 9 + 10 + 11 + 12 + 13 + 14 + 15 + 16$ Therefore, all possible ways 100 = 100 $100 = 18 + 19 + 20 + 21 + 22$ $100 = 9 + 10 + 11 + 12 + 13 + 14 + 15 + 16$

7. The number of bacteria in a certain culture doubles every hour. If there were 30 bacteria present in the culture originally, now many bacteria will be present at the end of the $2^{nd}$ hour, $4^{th}$ hour and $n^{th}$ hour?

Sol. Initial numbers of bacteria $= 30$ The bacteria double every hour at the end of $1^{st}$ hour $= 30 \times 2 = 60$ at the end of $2^{nd}$ hour $= 30 \times 2^{2}$ $= 30 \times 4 = 120$ at the end of $4^{th}$ hour $= 30 \times 2^{4}$ $= 30 \times 16 = 480$ at the end of $n^{th}$ hour $= 30 \times 2^{n}$ There At the end of $2^{th}$ hour $= 120$ bacteria At the end of $4^{th}$ hour $= 480$ bacteria At the end of $n^{th}$ hour $= 30 \times 2^{n}$ bacteria

8. The sum of the $4^{th}$ and $8^{th}$ terms of an AP is 24 and the sum of the $6^{th}$ and $1 0^{th}$ terms is 44 . Find the first three terms of the AP.

Sol. $t_{4} = a + 3 d, t_{8} = a + 7 d$ $a + 3 d + a + 7 d = 24$ $2 a + 10 d = 24$ $a + 5 d = 12$ $\\172.26.70.208\$ smd_data$\PNCF\2026-27\Print Module\SET-1\NCERT\Mathematics\9th $t_{6} = a + 5 d$ $t_{10} = a + 9 d$ So $t_{6} + t_{10} = 44$ $2 a + 14 d = 44$ $a + 7 d = 22$ equation (2) - (1) we get $2 d = 10$ $d = 5$ and $a + 5 (5) = 12$ $a = - 13$ First three terms are $= a, a + d, a + 2 d$ $= - 13, - 8, - 3$

9. Find the smallest value of n such that the sum of the first n natural numbers is greater than 1,000 .

Sol. Sum of first ' n ' natural numbers $S_{n} = \frac{n ( n + 1 )}{2}$ $\frac{n ( n + 1 )}{2} > 1000$ $\Rightarrow n (n + 1) > 2000$ Now checking $44 \times 45 = 1980$ So, $S_{44} = 990$ $45 \times 46 = 2070$ So, $= 480$ Since 1035 > 1000 Therefore, smallest value of n is 45 .

10. Which term of the GP: $2, 8, 32, \dots\dots$ is 131072? Write the explicit formula as well as the recursive formula for the $n^{th}$ term.

Sol. G.P. $= 2, 8, 32 \dots\dots$ Here $a = 2$ $r = \frac{8}{4} = 4$ ; since $t_{n} = ar^{n - 1}$ New $2 \times 4^{n - 1} = 131072$ $4^{n - 1} = 65536 = 4^{8}$ $n - 1 = 8$ $\Rightarrow n = 9$ Recursive formula $t_{n} = 4. t_{n - 1}$ for $n \geq 2$

11. The sum of the first three of a GP is $\frac{13}{12}$ and their product is -1 . Find the common ratio and the terms.

Sol. Let three terms are $= \frac{a}{r}, a, ar$

$Product = \frac{a}{r} \times a \times a r = a^{3} Product = a^{3} = - 1 \Rightarrow a = - 1$

So, terms $= \frac{- 1}{r}, - 1, - r$ Sum $= \frac{- 1}{r} - 1 - r = \frac{13}{12}$ $- 12 r^{2} - 12 r - 12 = 13 r$ $12 r^{2} + 25 r + 12 = 0$ $12 r^{2} + 16 r + 9 r + 12 = 0$ $4 r (3 r + 4) + 3 (3 r + 4) = 0$ $(4 r + 3) (3 r + 4) = 0$ So $r = \frac{- 3}{4}$ or $r = \frac{- 4}{3}$ If $r = \frac{- 3}{4}$ Terms are $= \frac{4}{3}, - 1, \frac{3}{4}$ If $r = \frac{- 4}{3}$ Terms are $= \frac{3}{4}, - 1, \frac{4}{3}$ Therefore the common ratio is $\frac{- 3}{4}$ or $\frac{- 4}{3}$ and the terms are $\frac{4}{3}, - 1, \frac{3}{4}$ or $\frac{3}{4}, - 1, \frac{4}{3}$

12. If the $4^{th}, 1 0^{th}$ and $1 6^{th}$ terms of a GP are x , y and z respectively, prove that $x, y, z$ are in G.P.

Sol. Let first term of GP is a and common ratio be r Then $4^{th}$ term $= ar^{3} = x$ $1 0^{th}$ term $= ar^{9} = y$ $16^{th}$ term $= ar^{15} = x$ Now $\frac{y}{x} = \frac{ar ^{9}}{ar ^{3}} = r^{6}$ also $\frac{z}{y} = \frac{ar ^{15}}{ar ^{9}} = r^{6}$ Therefore $\frac{y}{x} = \frac{z}{y} & y^{2} = xz$ So, $x, y, z$ in G.P.

13. The sum of the first three terms of a geometric progression is 26 , and the sum of their squares is 364 . Find the terms of the GP.

Sol. Let the three terms be a , ar , $ar^{2}$ According to first condition $a + ar + ar^{2} = 26$ $a (1 + r + r^{2}) = 26$ also $(a)^{2} + (ar)^{2} + (ar)^{2} = 364$ $a^{2} (1 + r^{2} + r^{4}) = 364$

Now $\frac{eq.(2)}{[ eq. (1) ] ^{2}}$ we have $\frac{a ^{2} ( 1 + r ^{2} + r ^{4} )}{[ a ( 1 + r + r ^{2} ) ] ^{2}} = \frac{364}{( 26 ) ^{2}}$ $\frac{1 + r ^{2} + r ^{4}}{( 1 + r + r ^{2} ) ^{2}} = \frac{7}{13}$ $13 (1 + r^{2} + r^{4}) = 7 (1 + r + r^{2})^{2}$ $13 (1 + r^{2} + r^{4}) - 7 (r^{2} + r + 1)^{2} = 0$ $13 (r^{2} + r + 1) (r^{2} - r + 1) - 7 (r^{2} + r + 1)^{2} = 0$

Using $1 + r^{2} + r^{4} = (r^{2} + r + 1) (r^{2} - r + 1)$ $13 (r^{2} - r + 1) - 7 (r^{2} + r + 1) = 0$ {Dividing by ( $r^{2} + r + r^{1}$ )} $[13 r^{2} - 13 r + 13 - 7 r^{2} - 7 r - 7] = 0$ $(6 r^{2} - 20 r + 6) = 0$ $3 r^{3} - 10 r + 3 = 0$ $(3 r - 1) (r - 3) = 0$ $r = \frac{1}{3}, r = 3$ If $r = \frac{1}{3}; a (1 + \frac{1}{3} + \frac{1}{3 ^{2}}) = 26$ $a \frac{( 9 + 3 + 1 )}{9} = 26$ $a = 18$ Hence the three terms a, ar, ar $^{2}$ are 18, 6, 2. If $r = 3$ from eq. (1) we have $a (1 + 3 + 3^{2}) = 26$ $a (1 + 3 + 9) = 26$ $a (13) = 26$ $a = 2$ So, three terms are $2, 6, 18$ .

14. Suppose $P_{1} = 1, P_{2} = 2$ and for $n > 2$ , $P_{n} = P_{1} + P_{2} + \dots\dots P_{n - 1} + 1$ . Find the values of $P_{1}, P_{2}, \dots\dots P_{8}$ . Can you find a simpler recursive formula for $P_{n}$ ? Can you give an explicit formula?

Sol. Given $P_{1} = 1, P_{2} = 2$ For $n > 2; P_{n} = P_{1} + P_{2} + \dots\dots + P_{n - 1} + 1$ Now $P_{3} = P_{1} + P_{2} + 1$ $P_{3} = 1 + 2 + 1 = 4$ $P_{3} = 4$ $P_{4} = P_{1} + P_{2} + P_{3} + 1$ $P_{4} = 1 + 2 + 4 + 1 = 8$ $P_{5} = 1 + 2 + 4 + 8 + 1 = 16$ $P_{6} = 1 + 2 + 4 + 8 + 16 + 1 = 32$ $P_{7} = 64$ $P_{8} = 128$ So $P_{1}, P_{2}, P_{3} \dots\dots .. P_{8}$ are $1, 2, 4, 8, 16, 32, 64$ , 128

Simpler recursive formula $P_{1} = 1$ $P_{n} = 2 P_{n - 1}$ for $n \geq 2$ Explicit formula $P_{n} = 2^{n - 1}$

15. Suppose $W_{1} = 1, W_{2} = 2$ and for $n > 2$ , $W_{n} = W_{1} + W_{2} + \dots\dots + W_{n - 2} + 2$ . Find the values of $W_{1}, W_{2}, + \dots\dots + W_{8}$ . Do you recognise this sequence?

Sol. Given $W_{1} = 1, W_{2} = 2$ For $n > 2$ $W_{n} = W_{1} + W_{2} + \dots\dots W_{n - 2} + 2$ $W_{3} = W_{1} + 2 = 1 + 2 = 3$ $W_{4} = W_{1} + W_{2} + 2 = 1 + 2 + 2 = 5$ $W_{5} = W_{1} + W_{2} + W_{3} + 2 = 1 + 2 + 3 + 2 = 8$ $W_{6} = W_{1} + W_{2} + W_{3} + W_{4} + 2$ $W_{6} = 1 + 2 + 3 + 5 + 2 = 13$ $W_{7} = 1 + 2 + 3 + 5 + 8 + 2 = 21$ $W_{8} = 1 + 2 + 3 + 5 + 8 + 13 + 2 = 34$ Therefore $W_{1}, W_{2} \dots\dots\dots .. W_{8}$ are $1, 2, 3, 5$ , 8, 13, 21, 34 Yes This is the virahanka - Fibonacci sequence.

4.0Important Key Concepts Class 9 Maths Exploring Algebraic Identities - Chapter 4

Topic	What Students Learn
Visualising (a + b)² and (a − b)²	A geometric derivation of these identities using squares and rectangles, building an intuitive picture of why the formulas hold before they're ever applied numerically
Expanding Algebraic Expressions	Step-by-step application of standard identities to expand expressions containing variables, fractions, and decimal coefficients accurately
Fast Numerical Calculations	Techniques to compute squares of large numbers, such as 117² or 78², by breaking them into convenient sums or differences instead of relying on long multiplication
Factorisation Using Algebra Tiles	A hands-on, visual method of factorising quadratic expressions by physically arranging tiles that represent each term
Factorisation Without Algebra Tiles	The splitting-the-middle-term method, which factorises quadratic expressions purely through algebraic manipulation
More Identities: (a + b + c)²	Expansion of the three-term square identity and its use in simplifying more complex expressions involving three variables
Sum and Difference of Cubes	The identities for a³ + b³ and a³ − b³, along with the special three-variable identity x³ + y³ + z³ − 3xyz, and how each applies to factorisation problems
Finding New Identities	Derivation of further identities such as (a + b)³, (a − b)³, and x³ − y³ = (x − y)(x² + xy + y²) by logically extending known identities rather than memorising them
Simplifying Rational Expressions	Combining factorisation with identity-based techniques to cancel common factors and reduce algebraic fractions to their simplest form

5.0Exercise-wise NCERT Class 9 Maths Chapter 4 Exploring Algebraic Identities

Chapter 4 of the newly updated NCERT syllabus is divided into five exercises and an End-of-Chapter set, moving students from basic identities to advanced factorisation. Here's what each exercise covers.

Exercise	Topics Covered
Exercise 4.1	Introduction to algebraic identities and their geometric interpretation using algebra tiles and area models.
Exercise 4.2	Verification and application of standard algebraic identities to expand algebraic expressions.
Exercise 4.3	Factorisation of algebraic expressions using algebraic identities and simplification of expressions.
Exercise 4.4	Discovering additional algebraic identities and applying them to solve mathematical problems.
End-of-Chapter Exercises	Mixed questions covering algebraic identities, factorisation, simplification, and application-based problems from the chapter

6.0Quick Revision on Class 9 Maths Chapter 4 - Exploring Algebraic Identities

Quick Rev cl9 maths ch4

7.0Related Study Materials Class 9 Maths

Continue your learning with ALLEN’s Class 9 Maths study material as per latest NCERT syllabus. Along with NCERT Solutions, NCERT textbooks, revision notes, sample papers and previous years question papers to strengthen your concepts, revise effectively and prepare with confidence for school and CBSE examinations.

CBSE Class 9 Maths Syllabus

Class 9 Maths Revision Notes

NCERT Textbook for Class 9 Maths

CBSE Sample Papers for Class 9 Maths

8.0Advantages of Chapter 4 Maths Class 9 NCERT Solutions

Develops understanding of algebraic expressions Clearly defines variables, constants, terms and coefficients.
Simplifies Algebraic Operations: It helps students in adding, subtracting, multiplying and dividing algebraic expressions correctly.
Develops Factorisation Skills Enhances the skill of factorising algebraic expressions through different methods.
Improves Expression Simplification: Teaches systematic ways to simplify complex algebraic expressions.
Increases Conceptual Clarity : Helps the students to identify different types of algebraic expressions and their components.
Provides a Solid Base in Algebra: Prepares students for solving algebraic problems and understanding difficult algebraic concepts in later grades.