Class 9 · Advanced MathematicsNCERT 2026 Revised Syllabus
Chapter 6: Exploring Some More Progressions
Class IX CBSE Advanced Mathematics
Extends progressions with geometric series, sums to infinity, repeated decimals, and special sequences where the second differences form an arithmetic progression, using the general $t_n$ and $S_n$ formulas.
Find complete NCERT solutions for Class 9 Advanced Mathematics, Class IX CBSE Advanced Mathematics Chapter 6. Every question has a clear, step-by-step answer.
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Exercise 6.1(13 questions)
1
HardFind the sum of
Find the sum of $0.15 + 0.015 + 0.0015 + \dots$ to 15 terms.
School Level Solution
Sol.
Solution:This is a GP with
$a = 0.15$, $r = 0.1$.$$S_{15} = \frac{a(1 - r^{15})}{1 - r} = \frac{0.15(1 - 0.1^{15})}{0.9}$$
$$= \frac{1}{6}\left(1 - 10^{-15}\right)$$Answer: $\boxed{S_{15} = \frac{1}{6}(1 - 10^{-15})}$.2
HardFind the sum
Find the sum $0.9 + 0.99 + 0.999 + \dots$ to $n$ terms.
School Level Solution
Sol.
Solution:
$$0.9 + 0.99 + 0.999 + \dots = (1 - 0.1) + (1 - 0.01) + (1 - 0.001) + \dots$$
$$= n - (0.1 + 0.01 + 0.001 + \dots + 0.1^n)$$The subtracted GP has $a = 0.1$, $r = 0.1$:
$$\text{sum} = \frac{0.1(1 - 0.1^n)}{0.9} = \frac{1 - 10^{-n}}{9}$$$$S_n = n - \frac{1 - 10^{-n}}{9}$$Answer: $\boxed{S_n = n - \frac{1 - 10^{-n}}{9}}$.3
HardFind the sum
Find the sum $5 + 55 + 555 + \dots$ to $n$ terms.
School Level Solution
Sol.
Solution:
$$S_n = 5(1 + 11 + 111 + \dots) = \frac{5}{9}(9 + 99 + 999 + \dots)$$
$$= \frac{5}{9}[(10 - 1) + (100 - 1) + (1000 - 1) + \dots]$$
$$= \frac{5}{9}[(10 + 100 + 1000 + \dots \text{ to } n \text{ terms}) - n]$$$$S_n = \frac{5}{9}\left[\frac{10(10^n - 1)}{9} - n\right]$$Answer: $\boxed{S_n = \frac{5}{9}\left[\frac{10(10^n - 1)}{9} - n\right]}$.4
HardThe GP
The GP $3, 6, 12, \dots$ has sum $381$. Find $n$.
School Level Solution
Sol.
Solution:
$a = 3$, $r = 2$:
$$S_n = \frac{3(2^n - 1)}{2 - 1} = 3(2^n - 1) = 381 \Rightarrow 2^n - 1 = 127 \Rightarrow 2^n = 128 = 2^7$$$$n = 7$$Answer: $\boxed{n = 7}$.5
HardFind the sum
Find the sum $x(x+y) + x^2(x^2 + y^2) + x^3(x^3 + y^3) + \dots$ to $n$ terms.
School Level Solution
Sol.
Solution:
$$= (x^2 + x^4 + x^6 + \dots + x^{2n}) + (xy + x^2y^2 + x^3y^3 + \dots + x^ny^n)$$First part is a GP: $a = x^2$, $r = x^2$. Second part is a GP: $a = xy$, $r = xy$.$$S_n = \frac{x^2(x^{2n} - 1)}{x^2 - 1} + \frac{xy[(xy)^n - 1]}{xy - 1}$$Answer: $\boxed{S_n = \frac{x^2(x^{2n} - 1)}{x^2 - 1} + \frac{xy[(xy)^n - 1]}{xy - 1}}$.6
HardFind the sum
Find the sum $1 + \frac{1}{2} + \frac{1}{9} + \frac{1}{4} + \frac{1}{9^2} + \frac{1}{8} + \frac{1}{9^3} + \dots$
School Level Solution
Sol.
Solution:Group the powers of 2 and powers of 9 separately:
$$\left(1 + \frac{1}{2} + \frac{1}{2^2} + \dots\right) + \left(\frac{1}{9} + \frac{1}{9^2} + \frac{1}{9^3} + \dots\right)$$First GP: $a = 1$, $r = \frac{1}{2}$, sum $= \frac{1}{1 - 1/2} = 2$.Second GP: $a = \frac{1}{9}$, $r = \frac{1}{9}$, sum $= \frac{1/9}{1 - 1/9} = \frac{1}{8}$.$$\text{Total} = 2 + \frac{1}{8} = \frac{17}{8}$$Answer: $\boxed{\frac{17}{8}}$.7
HardAn infinite GP has
An infinite GP has $\frac{a}{1 - r} = 6$ and $\frac{a^2}{1 - r^2} = 12$. Find $r$.
School Level Solution
Sol.
Solution:From (1):
$a = 6(1 - r) \Rightarrow a^2 = 36(1 - r)^2$.Substitute in (2):
$$\frac{36(1 - r)^2}{(1 - r)(1 + r)} = 12 \Rightarrow \frac{36(1 - r)}{1 + r} = 12 \Rightarrow 3(1 - r) = 1 + r$$
$$3 - 3r = 1 + r \Rightarrow 2 = 4r \Rightarrow r = \frac{1}{2}$$Answer: $\boxed{r = \frac{1}{2}}$.8
HardFor an infinite GP with
For an infinite GP with $S = \frac{a}{1 - r}$ where $a = 1$, express $r$ in terms of $S$.
School Level Solution
Sol.
Solution:With
$a = 1$:
$$S = \frac{1}{1 - r} \Rightarrow 1 - r = \frac{1}{S} \Rightarrow r = 1 - \frac{1}{S} = \frac{S - 1}{S}$$Answer: $\boxed{r = \frac{S - 1}{S}}$.9
HardFind the product
Find the product $4^{1/2} \times 4^{1/4} \times 4^{1/8} \times 4^{1/16} \times \dots$
School Level Solution
Sol.
Solution:Sum of exponents:
$$\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \dots = \frac{1/2}{1 - 1/2} = 1 \quad (\text{infinite GP})$$$$\text{Product} = 4^1 = 4$$Answer: $\boxed{\text{4}}$.10
Hard
An equilateral triangle has perimeter 48 cm. A second triangle is drawn by joining the midpoints of its sides (perimeter halved), and so on. Find the sum of perimeters of all such triangles to infinity.
School Level Solution
Sol.
Solution:Perimeter of
$\triangle ABC = 3 \times 16 = 48$. Each successive triangle\u2019s perimeter is half the previous: GP $48, 24, 12, \dots$ with $a = 48$, $r = \frac{1}{2}$.$$\text{Sum to infinity} = \frac{48}{1 - \frac{1}{2}} = 96$$Answer: $\boxed{\text{96 cm}}$.11
HardFor
For $f(x) = 2x + 1$, for what value of $x$ are $f(x)$, $f(2x)$ and $f(4x)$ in a GP?
School Level Solution
Sol.
Solution:
$f(x) = 2x + 1$, $f(2x) = 4x + 1$, $f(4x) = 8x + 1$.For a GP: $[f(2x)]^2 = f(x) \cdot f(4x)$:
$$(4x + 1)^2 = (2x + 1)(8x + 1)$$
$$16x^2 + 8x + 1 = 16x^2 + 10x + 1 \Rightarrow 8x = 10x \Rightarrow x = 0$$At $x = 0$, all three terms equal 1 — a valid GP with common ratio 1.Answer: $\boxed{\text{Only 1 value: }x = 0}$.12
HardIf
If $t_1, t_2, t_3, \dots$ is a GP with ratio $r$, $-1 < r < 1$, find $\frac{t_1 - t_3 + t_5 - \dots}{t_2 - t_4 + t_6 - \dots}$.
School Level Solution
Sol.
Solution:Numerator:
$$t_1 - t_3 + t_5 - \dots = t_1(1 - r^2 + r^4 - \dots) = \frac{t_1}{1 + r^2}$$Denominator:
$$t_2 - t_4 + t_6 - \dots = t_2(1 - r^2 + r^4 - \dots) = \frac{t_2}{1 + r^2}$$$$\frac{t_1 - t_3 + t_5 - \dots}{t_2 - t_4 + t_6 - \dots} = \frac{t_1}{t_2} = \frac{t_1}{t_1 r} = \frac{1}{r}$$Answer: $\boxed{\frac{1}{r}}$.13
HardA ball is dropped from height 90 m; on each bounce it rebounds to
A ball is dropped from height 90 m; on each bounce it rebounds to $\frac{3}{5}$ of its previous height. Find the total distance travelled until it comes to rest.
School Level Solution
Sol.
Solution:
$$\text{Total distance} = h + 2hr + 2hr^2 + \dots = h + \frac{2hr}{1 - r}$$
$$= 90 + \frac{2(90)\left(\frac{3}{5}\right)}{1 - \frac{3}{5}} = 90 + \frac{108}{\frac{2}{5}} = 90 + 270 = 360$$Answer: $\boxed{\text{360 metres}}$.Exercise 6.2(4 questions)
1
HardFor the series
For the series $1, 9, 24, 46, 75, \dots$, find the general term $t_n$ and the sum $S_n$.
School Level Solution
Sol.
Solution:First differences (Row 2):
$8, 15, 22, 29$ — an AP with first term 8, common difference 7.So $b = 1$, $a = 8$, $d = 7$.$$t_n = b + a(n-1) + d\frac{(n-1)(n-2)}{2} = 1 + 8(n-1) + \frac{7}{2}(n-1)(n-2)$$On simplification:
$$t_n = \frac{7n^2 - 5n}{2} = \frac{n(7n - 5)}{2}$$Check: $n = 1: 1; n = 2: 9; n = 3: 24; n = 4: 46; n = 5: 75$ — all match.$$S_n = bn + a\frac{n(n-1)}{2} + d\frac{n(n-1)(n-2)}{6}$$On simplification:
$$S_n = \frac{n(n+1)(7n - 4)}{6}$$Answer: $\boxed{t_n = \frac{n(7n - 5)}{2}\text{, }S_n = \frac{n(n+1)(7n - 4)}{6}}$.2
HardFor the series
For the series $4, 5, 9, 16, 26, \dots$, find $t_{10}$ and $S_{10}$.
School Level Solution
Sol.
Solution:First differences:
$1, 4, 7, 10$ — AP with first term 1, common difference 3.So $b = 4$, $a = 1$, $d = 3$.$$t_n = 4 + (n-1) + \frac{3}{2}(n-1)(n-2)$$$$t_{10} = 4 + 9 + \frac{3}{2}(9)(8) = 4 + 9 + 108 = 121$$$$S_{10} = b(10) + a\frac{10 \times 9}{2} + d\frac{10 \times 9 \times 8}{6} = 4(10) + 1(45) + 3(120) = 40 + 45 + 360 = 445$$Answer: $\boxed{t_{10} = 121\text{, }S_{10} = 445}$.3
HardFor the series
For the series $3, 6, 11, 18, 27, \dots$, find the general term $t_n$ and $S_{12}$.
School Level Solution
Sol.
Solution:First differences:
$3, 5, 7, 9$ — AP with first term 3, common difference 2.So $b = 3$, $a = 3$, $d = 2$.$$t_n = 3 + 3(n-1) + (n-1)(n-2)$$On simplification:
$$t_n = n^2 + 2$$Check: $n = 1: 3; n = 2: 6; n = 3: 11; n = 4: 18; n = 5: 27$ — all match.$$S_{12} = \sum_{k=1}^{12}(k^2 + 2) = \frac{12 \times 13 \times 25}{6} + 2(12) = 650 + 24 = 674$$Answer: $\boxed{t_n = n^2 + 2\text{, }S_{12} = 674}$.4
HardFor the series
For the series $4, 13, 28, 49, 76, \dots$, find $t_8$ and $S_8$.
School Level Solution
Sol.
Solution:First differences:
$9, 15, 21, 27$ — AP with first term 9, common difference 6.So $b = 4$, $a = 9$, $d = 6$.$$t_n = 4 + 9(n-1) + 3(n-1)(n-2)$$$$t_8 = 4 + 9(7) + 3(7)(6) = 4 + 63 + 126 = 193$$$$S_8 = b(8) + a\frac{8 \times 7}{2} + d\frac{8 \times 7 \times 6}{6} = 4(8) + 9(28) + 6(56) = 32 + 252 + 336 = 620$$Answer: $\boxed{t_8 = 193\text{, }S_8 = 620}$.