Class 11 Maths · Chapter 8
Short answer:
Class 11 Maths Chapter 8 Sequences and Series NCERT solutions — Arithmetic Progression (AP), Geometric Progression (GP), AM-GM relation, aur special series (Σn, Σn², Σn³) ke solved intext questions, exercise questions, key formulas, common mistakes aur important practice questions, step-by-step working ke saath.
Class 11 Maths Chapter 8 — Sequences and Series — wahi chapter hai jahan pehli baar 'pattern' ko formally define kiya jaata hai: number ki koi bhi ordered list ek sequence hoti hai, aur jab us list ke terms ko '+' se jod diya jaaye to wo ek series ban jaati hai. Yahin se do sabse important patterns nikalte hain — Arithmetic Progression (AP, jahan consecutive terms ka difference constant rehta hai) aur Geometric Progression (GP, jahan consecutive terms ka ratio constant rehta hai).
Ye chapter sirf apne aap me important nahi hai — iske formulas (nth term, sum of n terms, AM-GM relation, special series Σn/Σn²/Σn³) aage Binomial Theorem aur Limits and Derivatives me bhi direct use hote hain, isliye is chapter ko skip karna risky hota hai. Agar tum class 11 maths ncert solutions chapter-by-chapter dekh rahe ho, to Chapter 1 (Sets) aur Chapter 3 (Trigonometric Functions) ke baad Sequences and Series algebra ka teesra bada pillar hai — aur is naye rationalised syllabus (session 2026-27) me numbering thodi shift hui hai kyunki Mathematical Induction aur Mathematical Reasoning wale chapters hata diye gaye hain, isliye apna class 11 maths syllabus 2026-27 pdf ya ncert book pdf download karke chapter list ek baar zaroor confirm kar lo.
Neeche solved intext questions (sequences ke nth-term problems), representative exercise questions (AP, GP, special series aur miscellaneous se), key formulas table, common mistakes, aur exam-style practice questions diye gaye hain — sab clear step-by-step working ke saath.
Chapter 8 Summary — 5 Minute Revision
Class 11 Maths Chapter 8 Sequences and Series ek line me: sequence = ordered list of numbers following a rule, series = us list ka sum. Do central patterns hain — A.P. (constant common difference d, nth term a+(n-1)d, sum n/2[2a+(n-1)d]) aur G.P. (constant common ratio r, nth term a·r^(n-1), finite sum a(rⁿ-1)/(r-1), aur |r| < 1 ho to infinite sum a/(1-r)).
Do numbers a, b ke beech Arithmetic Mean A = (a+b)/2 aur Geometric Mean G = √(ab) hoti hai, aur inke beech hamesha A ≥ G relation rehta hai (equality sirf a=b pe). Chapter ke end me teen special series ke ready-made sum formulas milte hain — Σn = n(n+1)/2, Σn² = n(n+1)(2n+1)/6, aur Σn³ = [n(n+1)/2]² — jo complex nth-term series (jaise n(n+1), n²+2ⁿ) ko split karke solve karne me kaam aate hain. Ye poora toolkit — AP, GP, AM-GM, aur special series — aage Binomial Theorem aur Limits/Derivatives me directly reuse hota hai.
In-Text Questions — Solutions
Write the first five terms of the sequence whose nth term is a_n = n(n+2).
Har term k = 1,2,3,4,5 rakh kar formula me substitute karte hain:
a_1 = 1(3) = 3
a_2 = 2(4) = 8
a_3 = 3(5) = 15
a_4 = 4(6) = 24
a_5 = 5(7) = 35
Isliye first five terms: 3, 8, 15, 24, 35
Write the first five terms of the sequence whose nth term is a_n = n/(n+1).
a_1 = 1/2
a_2 = 2/3
a_3 = 3/4
a_4 = 4/5
a_5 = 5/6
Sequence: 1/2, 2/3, 3/4, 4/5, 5/6
Write the first five terms of the sequence whose nth term is a_n = 2ⁿ.
a_1 = 2¹ = 2
a_2 = 2² = 4
a_3 = 2³ = 8
a_4 = 2⁴ = 16
a_5 = 2⁵ = 32
Sequence: 2, 4, 8, 16, 32
Write the first five terms of the sequence whose nth term is a_n = (2n-3)/6.
a_1 = (2-3)/6 = -1/6
a_2 = (4-3)/6 = 1/6
a_3 = (6-3)/6 = 1/2
a_4 = (8-3)/6 = 5/6
a_5 = (10-3)/6 = 7/6
Sequence: -1/6, 1/6, 1/2, 5/6, 7/6
Write the first five terms of the sequence whose nth term is a_n = (-1)^(n-1) · 5^(n+1).
Sign (-1)^(n-1) odd n pe positive aur even n pe negative deta hai:
a_1 = (+1)(5²) = 25
a_2 = (-1)(5³) = -125
a_3 = (+1)(5⁴) = 625
a_4 = (-1)(5⁵) = -3125
a_5 = (+1)(5⁶) = 15625
Sequence: 25, -125, 625, -3125, 15625
If a_n = 4n - 3, find a_17 and a_24.
a_17 = 4(17) - 3 = 68 - 3 = 65
a_24 = 4(24) - 3 = 96 - 3 = 93
Isliye a_17 = 65 aur a_24 = 93
If a_n = n²/2ⁿ, find a_7.
Direct substitution:
a_7 = 7²/2⁷ = 49/128
Isliye a_7 = 49/128
Let the Fibonacci sequence be defined as a_1 = a_2 = 1, a_n = a_{n-1} + a_{n-2} for n > 2. Find a_{n+1}/a_n for n = 1, 2, 3, 4, 5.
Pehle terms generate karte hain:
a_1=1, a_2=1, a_3=2, a_4=3, a_5=5, a_6=8
Ab ratios nikalte hain:
n=1: a_2/a_1 = 1/1 = 1
n=2: a_3/a_2 = 2/1 = 2
n=3: a_4/a_3 = 3/2
n=4: a_5/a_4 = 5/3
n=5: a_6/a_5 = 8/5
Isliye ratios: 1, 2, 3/2, 5/3, 8/5
A sequence is defined by a_1 = 3, a_n = 3a_{n-1} + 2 for all n > 1. Find the first five terms and the corresponding series.
Recursive formula ko baar-baar apply karte hain:
a_1 = 3
a_2 = 3(3)+2 = 11
a_3 = 3(11)+2 = 35
a_4 = 3(35)+2 = 107
a_5 = 3(107)+2 = 323
First five terms: 3, 11, 35, 107, 323
Corresponding series: 3 + 11 + 35 + 107 + 323 + ...
A sequence is defined by a_1 = -1, a_n = a_{n-1}/n for n ≥ 2. Find the first five terms.
a_1 = -1
a_2 = a_1/2 = -1/2
a_3 = a_2/3 = -1/6
a_4 = a_3/4 = -1/24
a_5 = a_4/5 = -1/120
First five terms: -1, -1/2, -1/6, -1/24, -1/120
Poore Class 11 Maths ke handwritten colour notes
IITian & district toppers ke banaye short notes — revision-ready, diagram ke saath. Board se pehle poora syllabus 3 din me revise.
Exercise Questions — Solutions (Q1–Q22)
(AP) Find the sum of odd integers from 1 to 2001.
Odd integers 1, 3, 5, ..., 2001 hain apne aap me ek AP jisme a = 1, d = 2, l (last term) = 2001.
Pehle n (number of terms) nikalte hain:
l = a + (n-1)d
2001 = 1 + (n-1)(2) ⟹ 2000 = 2(n-1) ⟹ n - 1 = 1000 ⟹ n = 1001
Ab sum formula lagate hain:
S_n = n/2 (a + l) = 1001/2 × (1 + 2001) = 1001/2 × 2002
S_n = 1001 × 1001 = 1002001
Shortcut: pehle n odd numbers ka sum hamesha n² hota hai, aur yahan 1001 odd numbers hain, isliye 1001² = 1002001 seedha bhi mil jaata hai.
(AP) Find the sum of all natural numbers between 100 and 1000 which are multiples of 5.
Multiples of 5 between 100 and 1000 (100 aur 1000 dono exclude, kyunki 'between' bola gaya hai): 105, 110, 115, ..., 995.
Yahan a = 105, d = 5, l = 995.
995 = 105 + (n-1)(5) ⟹ 890 = 5(n-1) ⟹ n - 1 = 178 ⟹ n = 179
S_n = n/2 (a+l) = 179/2 × (105 + 995) = 179/2 × 1100 = 179 × 550
S_n = 98450
(AP) If the sum of n terms of an A.P. is (pn + qn²), find its common difference.
Diya hai S_n = pn + qn². nth term nikalne ke liye:
a_n = S_n - S_{n-1}
a_n = (pn+qn²) - [p(n-1) + q(n-1)²]
a_n = pn + qn² - pn + p - q(n² - 2n + 1)
a_n = p - q + 2qn
Ye ek linear expression hai n me, isliye AP ka common difference n ka coefficient hota hai:
d = 2q
Check: a_1 = p+q, a_2 = p+3q ⟹ a_2 - a_1 = 2q. Match ho gaya.
(AP) The sum of first n terms of an A.P. is 4n - n². What is the first term, the sum of first two terms, the second term, and the nth term?
Diya S_n = 4n - n².
S_1 = 4(1) - 1² = 3 ⟹ pehla term a_1 = 3
S_2 = 4(2) - 2² = 4 ⟹ pehle do terms ka sum = 4
a_2 = S_2 - S_1 = 4 - 3 = 1
nth term ke liye general formula nikalte hain:
a_n = S_n - S_{n-1} = (4n-n²) - [4(n-1) - (n-1)²]
a_n = 4n - n² - 4n + 4 + (n² - 2n + 1) = 5 - 2n
Verify: a_1 = 5-2 = 3 ✓, a_2 = 5-4 = 1 ✓
Isliye a_1 = 3, S_2 = 4, a_2 = 1, a_n = 5 - 2n
(AP) The sums of n terms of two A.P.'s are in the ratio (5n+4):(9n+6). Find the ratio of their 18th terms.
Property: kisi bhi AP me a_m = S_{2m-1}/(2m-1), isliye do APs ke mth terms ka ratio, unke S_{2m-1} ka ratio hota hai.
Yahan m = 18 ⟹ n = 2m - 1 = 35 rakhna hoga diye gaye ratio expression me.
Ratio of 18th terms = (5(35)+4) / (9(35)+6) = (175+4)/(315+6) = 179/321
179 prime number hai aur 321 = 3 × 107, koi common factor nahi, isliye ratio simplest form me hi hai:
179 : 321
(AP) Between 1 and 31, m numbers have been inserted so that the resulting sequence is an A.P., and the ratio of the 7th and (m-1)th numbers is 5:9. Find m.
Poori sequence: 1, A₁, A₂, ..., A_m, 31 — total (m+2) terms, first term a=1, last term l=31.
d = (31-1)/(m+1) = 30/(m+1)
A₇ poori sequence ka 8th term hai, aur A_{m-1} poori sequence ka mth term hai:
A_7 = a + 7d = 1 + 7d, A_{m-1} = a + (m-1)d = 1 + (m-1)d
Diya hai (1+7d)/(1+(m-1)d) = 5/9:
9(1+7d) = 5(1+(m-1)d) ⟹ 9 + 63d = 5 + 5(m-1)d
4 = d[5(m-1) - 63] = d(5m - 68)
d = 30/(m+1) substitute karo:
4 = [30/(m+1)](5m-68) ⟹ 4(m+1) = 30(5m-68)
4m + 4 = 150m - 2040 ⟹ 2044 = 146m ⟹ m = 14
(GP) Find the 20th and nth terms of the G.P. 5/2, 5/4, 5/8, ....
Yahan a = 5/2, aur common ratio:
r = (5/4) ÷ (5/2) = 1/2
Formula: a_n = a·r^(n-1)
a_n = (5/2)(1/2)^(n-1) = 5/2ⁿ
n = 20 ke liye:
a_20 = 5/2²⁰
General nth term: a_n = 5/2ⁿ
(GP) Find the 12th term of a G.P. whose 8th term is 192 and the common ratio is 2.
Diya a_8 = 192, r = 2. Pehle first term a nikalte hain:
a_8 = a·r⁷ = a(2)⁷ = 128a = 192 ⟹ a = 192/128 = 3/2
Ab a_12 nikalte hain:
a_12 = a·r¹¹ = (3/2)(2)¹¹ = (3/2)(2048)
a_12 = 3072
(GP) The 5th, 8th and 11th terms of a G.P. are p, q and s respectively. Show that q² = ps.
Let first term A ho aur common ratio R.
p = A·R⁴, q = A·R⁷, s = A·R¹⁰
q² nikalte hain:
q² = A²R¹⁴
Ab ps nikalte hain:
ps = (A·R⁴)(A·R¹⁰) = A²R¹⁴
Dono equal hain, isliye:
q² = ps (proved)
(GP) Find the sum to n terms of the sequence 7, 77, 777, 7777, ....
Har term ko 7 common factor nikaal ke likhte hain:
S_n = 7[1 + 11 + 111 + ... n terms]
Andar wale expression ko 9 se multiply-divide karte hain taaki geometric pattern bane:
S_n = (7/9)[9 + 99 + 999 + ... n terms] = (7/9)[(10-1)+(100-1)+(1000-1)+...]
S_n = (7/9)[(10+10²+...+10ⁿ) - n]
Andar wala 10, 10², ..., 10ⁿ ek GP hai jiska sum = 10(10ⁿ-1)/9:
S_n = (7/9)[ 10(10ⁿ-1)/9 - n ]
S_n = (7/81)(10ⁿ⁺¹ - 10) - 7n/9
(GP) How many terms of the G.P. 3, 3/2, 3/4, ... are needed to give the sum 3069/512?
Yahan a = 3, r = 1/2 (kyunki r < 1 use hoga).
S_n = a(1-rⁿ)/(1-r) = 3(1-(1/2)ⁿ)/(1/2) = 6[1-(1/2)ⁿ]
Diya S_n = 3069/512, equation banate hain:
6 - 6/2ⁿ = 3069/512 ⟹ 6/2ⁿ = 6 - 3069/512 = (3072-3069)/512 = 3/512
2ⁿ = (6 × 512)/3 = 1024 = 2¹⁰
n = 10
(GP) The sum of first three terms of a G.P. is 39/10 and their product is 1. Find the common ratio and the terms.
Teen terms ko a/r, a, ar maan lete hain (symmetric form, GP ke beech wale term ke around).
Product: (a/r)(a)(ar) = a³ = 1 ⟹ a = 1
Sum: 1/r + 1 + r = 39/10 ⟹ 1/r + r = 29/10
Dono taraf r se multiply karte hain:
1 + r² = 29r/10 ⟹ 10r² - 29r + 10 = 0
Quadratic formula:
D = 841 - 400 = 441 = 21² ⟹ r = (29 ± 21)/20 = 5/2 ya 2/5
r = 5/2 lene par terms: a/r = 2/5, a = 1, ar = 5/2.
Terms: 2/5, 1, 5/2 (common ratio r = 5/2 ya 2/5, order ke hisaab se)
(GP) If the pth, qth and rth terms of a G.P. are a, b and c respectively, prove that a^(q-r) · b^(r-p) · c^(p-q) = 1.
Let first term A, common ratio R.
a = AR^(p-1), b = AR^(q-1), c = AR^(r-1)
Expression banate hain:
a^(q-r)·b^(r-p)·c^(p-q) = A^{(q-r)+(r-p)+(p-q)} · R^{(p-1)(q-r)+(q-1)(r-p)+(r-1)(p-q)}
A ka exponent: (q-r)+(r-p)+(p-q) = 0, isliye A-part = 1.
R ka exponent expand karte hain:
(p-1)(q-r)+(q-1)(r-p)+(r-1)(p-q) = [p(q-r)+q(r-p)+r(p-q)] - [(q-r)+(r-p)+(p-q)]
Pehla bracket identity se 0 hota hai (standard algebraic identity), doosra bracket bhi 0. Isliye R ka exponent bhi 0.
a^(q-r)·b^(r-p)·c^(p-q) = A⁰·R⁰ = 1 (proved)
(Special Series) Find the sum to n terms of the series 1×2 + 2×3 + 3×4 + 4×5 + ....
nth term: a_n = n(n+1) = n² + n
Sum ko do special series me tod dete hain:
S_n = Σn² + Σn = n(n+1)(2n+1)/6 + n(n+1)/2
n(n+1) common le lete hain:
S_n = n(n+1)[ (2n+1)/6 + 1/2 ] = n(n+1)[ (2n+1+3)/6 ]
S_n = n(n+1)(2n+4)/6 = n(n+1)×2(n+2)/6
S_n = n(n+1)(n+2)/3
(Special Series) Find the sum to n terms of the series whose nth term is n² + 2ⁿ.
Sum ko do parts me split karte hain — ek special series (Σn²) aur ek GP (Σ2ⁿ):
S_n = Σn² + Σ2ⁿ
Σn² = n(n+1)(2n+1)/6
GP part: a=2, r=2, isliye Σ2ⁿ (n terms) = 2(2ⁿ-1)/(2-1) = 2ⁿ⁺¹ - 2
S_n = n(n+1)(2n+1)/6 + 2ⁿ⁺¹ - 2
(Special Series) Find the sum to n terms of the series 3×1² + 5×2² + 7×3² + ....
Pattern dekho: kth term me coefficient (2k+1) hai aur square k² ka:
a_k = (2k+1)k² = 2k³ + k²
Sum karte hain:
S_n = 2Σk³ + Σk² = 2[n(n+1)/2]² + n(n+1)(2n+1)/6
S_n = n²(n+1)²/2 + n(n+1)(2n+1)/6
n(n+1) common le lete hain:
S_n = n(n+1)[ n(n+1)/2 + (2n+1)/6 ] = n(n+1)[ (3n(n+1)+2n+1)/6 ]
S_n = n(n+1)(3n²+5n+1)/6
S_n = n(n+1)(3n²+5n+1)/6
(Special Series) Find the sum to n terms of the series 1² + (1²+2²) + (1²+2²+3²) + ....
kth term khud ek sum hai:
t_k = 1²+2²+...+k² = k(k+1)(2k+1)/6 = (2k³+3k²+k)/6
Ab S_n = Σt_k, k=1 to n:
S_n = (1/6)[2Σk³ + 3Σk² + Σk]
S_n = (1/6)[ 2(n(n+1)/2)² + 3·n(n+1)(2n+1)/6 + n(n+1)/2 ]
n(n+1)/2 common nikaal lete hain:
S_n = (1/6)·[n(n+1)/2]·[n(n+1) + (2n+1) + 1]
S_n = [n(n+1)/12]·[n²+3n+2] = [n(n+1)/12]·(n+1)(n+2)
S_n = n(n+1)²(n+2)/12
(Miscellaneous) Show that the sum of the (m+n)th and (m-n)th terms of an A.P. is equal to twice the mth term.
Let AP ka first term a aur common difference d ho.
a_{m+n} = a + (m+n-1)d, a_{m-n} = a + (m-n-1)d
Dono ko add karte hain:
a_{m+n} + a_{m-n} = 2a + (2m-2)d = 2[a + (m-1)d]
= 2·a_m
Isliye a_{m+n} + a_{m-n} = 2a_m (proved)
(Miscellaneous) If A.M. and G.M. of two positive numbers a and b are A and G respectively, find the two numbers.
Diya hai A = (a+b)/2 aur G = √(ab), isliye a+b = 2A aur ab = G².
a aur b us quadratic equation ke roots hain jiska sum = a+b aur product = ab ho:
x² - (2A)x + G² = 0
Quadratic formula:
x = [2A ± √(4A²-4G²)]/2 = A ± √(A²-G²)
Isliye do numbers hain:
A + √(A²-G²) aur A - √(A²-G²)
(Miscellaneous) Find the sum to n terms of the series 0.6 + 0.66 + 0.666 + ....
6 common factor nikaalte hain:
S_n = 6[0.1 + 0.11 + 0.111 + ... n terms]
Andar 9 se multiply-divide karte hain:
S_n = (6/9)[0.9 + 0.99 + 0.999 + ...] = (2/3)[(1-0.1)+(1-0.01)+(1-0.001)+...]
S_n = (2/3)[ n - (0.1+0.01+...+0.1ⁿ) ]
Andar wala ek GP hai a=0.1, r=0.1, sum = 0.1(1-0.1ⁿ)/0.9 = (1/9)(1-10⁻ⁿ):
S_n = (2n/3) - (2/27)(1 - 10⁻ⁿ)
(Miscellaneous) Show that the sum of two numbers is 6 times their geometric mean, and hence show the numbers are in the ratio (3+2√2):(3-2√2).
Let numbers a aur b, diya hai a+b = 6√(ab).
Dono taraf b se divide karke k = √(a/b) rakhte hain:
(a/b) + 1 = 6√(a/b) ⟹ k² + 1 = 6k ⟹ k² - 6k + 1 = 0
k = [6 ± √(36-4)]/2 = [6 ± √32]/2 = 3 ± 2√2
Isliye a/b = k² = (3+2√2)² ya (3-2√2)². Ab identity check karte hain:
(3+2√2)/(3-2√2) — denominator ko rationalise karo (multiply by (3+2√2)):
= (3+2√2)² / (9-8) = (3+2√2)²
Ye k² ke barabar hai, isliye a:b confirm hota hai:
a : b = (3+2√2) : (3-2√2) (proved)
(Miscellaneous) If a, b, c, d are in G.P., show that (a²+b²+c²)(b²+c²+d²) = (ab+bc+cd)².
Since a,b,c,d GP me hain, first term A aur common ratio r lete hue: b=Ar, c=Ar², d=Ar³.
LHS nikalte hain:
a²+b²+c² = A²(1+r²+r⁴)
b²+c²+d² = A²r²(1+r²+r⁴)
LHS = A²(1+r²+r⁴) × A²r²(1+r²+r⁴) = A⁴r²(1+r²+r⁴)²
RHS nikalte hain:
ab+bc+cd = A²r + A²r³ + A²r⁵ = A²r(1+r²+r⁴)
RHS = [A²r(1+r²+r⁴)]² = A⁴r²(1+r²+r⁴)²
LHS = RHS, isliye (a²+b²+c²)(b²+c²+d²) = (ab+bc+cd)² (proved)
Important Equations — Ek Nazar Me
| Concept | Formula |
|---|---|
| A.P. — nth term | an = a + (n-1)d |
| A.P. — sum of n terms | Sn = n/2 [2a + (n-1)d] = n/2 (a + l) |
| Arithmetic Mean of a, b | A = (a+b)/2 |
| n Arithmetic Means between a, b | d = (b-a)/(n+1) |
| G.P. — nth term | an = a·rn-1 |
| G.P. — sum of n terms (r ≠ 1) | Sn = a(rn-1)/(r-1) for r > 1; Sn = a(1-rn)/(1-r) for r < 1 |
| G.P. — sum to infinity | S∞ = a/(1-r), valid only when |r| < 1 |
| Geometric Mean of a, b | G = √(ab) |
| AM–GM relation | A ≥ G, i.e. (a+b)/2 ≥ √(ab), equality iff a = b |
| Sum of first n natural numbers | Σn = n(n+1)/2 |
| Sum of squares of first n natural numbers | Σn² = n(n+1)(2n+1)/6 |
| Sum of cubes of first n natural numbers | Σn³ = [n(n+1)/2]² |
↔ Table ko side me swipe karein
Common Mistakes — Yahan Marks Kat te Hain
- AP aur GP ke pattern-check formula ko mix kar dena — d = a₂ - a₁ (subtract) AP ke liye hai, r = a₂/a₁ (divide) GP ke liye; galti se dono jagah subtract ya divide karna, khaaskar jab terms negative ho.
- GP ke sum formula me sign confuse karna — S_n = a(rⁿ-1)/(r-1) tab use karo jab r > 1, aur S_n = a(1-rⁿ)/(1-r) tab jab r < 1; value same hoti hai par galat wala formula use karne se calculation me sign error aa jaata hai.
- Sum to infinity ka formula S_∞ = a/(1-r) bina |r| < 1 check kiye laga dena — agar r ≥ 1 hai to series diverge karti hai aur infinite sum exist hi nahi karta, is condition ko bhoolna sabse common galti hai.
- Jab S_n diya ho to nth term nikalne ke liye a_n = S_n - S_{n-1} use karna bhool jaana, ya n=1 ke special case (a_1 = S_1, kyunki S_0 exist nahi karta) me galti karna.
- 'Insert n arithmetic/geometric means between a and b' wale problems me total terms ki counting galat karna — total terms (n+2) hote hain, isliye d = (b-a)/(n+1) hota hai, na ki (b-a)/n.
- Special series formulas Σn², Σn³ ko aapas me confuse karna — Σn³ = [n(n+1)/2]² (square wala) aur Σn² = n(n+1)(2n+1)/6 (alag formula), inhe ulta yaad rakh lena bahut common mistake hai.
Board-Style Important Questions
- Practice Q1: If the nth term of an A.P. is a_n = 3n - 7, write its first term and common difference.
- Practice Q2: Find the 10th term of the G.P. 3, 6, 12, 24, ....
- Practice Q3: The sum of the first three terms of a G.P. is 39/10 and their product is 1. Find the common ratio and the three terms.
- Practice Q4: If the pth, qth and rth terms of an A.P. are a, b, c respectively, show that a(q-r) + b(r-p) + c(p-q) = 0.
- Practice Q5: Find the sum to n terms of the series 3×1² + 5×2² + 7×3² + ..., and use it to find the sum for n = 5.
- Practice Q6: Insert three arithmetic means between 3 and 19, and find their sum.
Aksar Poochhe Jaane Wale Sawaal
Class 11 Maths Chapter 8 me sequence aur series me kya fark hai?
Sequence numbers ki ek ordered list hai jo ek definite pattern (rule) follow karti hai — jaise 2, 4, 6, 8, .... Jab in numbers ko '+' se jod diya jaata hai (2+4+6+8+...), tab wo series ban jaata hai. Simply: sequence = list, series = us list ka sum.
Sequences and Series ke important formulas kya hain — kya inki PDF milti hai?
Is chapter ke core formulas hain — AP ka nth term aur sum, GP ka nth term aur sum (including sum to infinity), AM-GM relation, aur special series (Σn, Σn², Σn³). Ye is page ke 'formulas' table me diye gaye hain; agar class 11 maths formulas pdf download chahiye poore syllabus ke liye, to har chapter ke formula table ko combine karke apna revision sheet bana sakte ho.
NCERT Class 11 Maths Sequences and Series me kitne exercises aur kitne questions hain?
Chapter me chaar numbered exercises hain — Sequences (8.1), Arithmetic Progression (8.2), Geometric Progression (8.3), aur Special Series (8.4) — plus ek Miscellaneous Exercise. Ye chapter GP wale exercise ki wajah se apne aap me lambi hai; is page pe har section se representative solved questions diye gaye hain, poore Ex 8.1–8.4 aur misc ka spread cover karte hue.
Kya Sequences and Series board exam ke liye important chapter hai?
Haan — ye chapter algebra ka core building block hai aur agle chapters (Binomial Theorem, Limits and Derivatives) me repeatedly use hota hai. Class 11 maths chapter 8 sequences and series important questions me AP-GP mix problems, AM≥GM proofs, aur special-series sums sabse zyada aate hain — is page ke exercise aur pyq section me wahi pattern cover kiya gaya hai.
Class 11 Maths ke baaki chapters ke NCERT solutions kahan milenge?
Yahi structure — worked examples, exercises, formulas, common mistakes — baaki chapters pe bhi follow hota hai: class 11 maths chapter 1 sets ncert solutions, class 11 maths chapter 3 trigonometric functions ncert solutions, class 11 maths straight lines ncert solutions, class 11 maths limits and derivatives ncert solutions, aur conic sections, probability jaise later chapters bhi. Poore class 11 maths ncert solutions set ke liye chapter-wise page dekho, aur exemplar-level practice ke liye class 11 maths exemplar solutions bhi try karo.
AM aur GM me relation kya hai — kya AM hamesha GM se bada hota hai?
Do positive numbers a aur b ke liye Arithmetic Mean A = (a+b)/2 aur Geometric Mean G = √(ab) hoti hai. Inke beech relation hai A ≥ G, yaani (a+b)/2 ≥ √(ab), aur equality tabhi hoti hai jab a = b ho. Ye result 'sum of a positive number and its reciprocal is ≥ 2' jaisi inequalities prove karne me kaam aata hai.
Class 11 Maths — Saare Chapters
Board exam tak sirf revision karna hai?
Class 11 Maths ke saare chapters ke colour handwritten short notes — diagrams, formulas aur important points ek jagah.
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