Write the first five terms of the sequence whose nth term is an = n(n+2).
Hint. Substitute n=1,2,3,4,5 directly into the formula.
a1=1(3)=3, a2=2(4)=8, a3=3(5)=15, a4=4(6)=24, a5=5(7)=35.
✦ Working through each part gives: 3, 8, 15, 24, 35.

Write the first five terms of the sequence whose nth term is an = n(n+2).
Hint. Substitute n=1,2,3,4,5 directly into the formula.
a1=1(3)=3, a2=2(4)=8, a3=3(5)=15, a4=4(6)=24, a5=5(7)=35.
✦ Working through each part gives: 3, 8, 15, 24, 35.
Write the first five terms of the sequence whose nth term is an = n/(n+1).
Hint. Substitute n=1,2,3,4,5 directly into the formula.
a1=1/2, a2=2/3, a3=3/4, a4=4/5, a5=5/6.
✦ Working through each part gives: 1/2, 2/3, 3/4, 4/5, 5/6.
Write the first five terms of the sequence whose nth term is an = 2^n.
Hint. Substitute n=1,2,3,4,5 directly into the formula.
a1=2^1=2, a2=2^2=4, a3=2^3=8, a4=2^4=16, a5=2^5=32.
✦ Working through each part gives: 2, 4, 8, 16, 32.
Write the first five terms of the sequence whose nth term is an = (2n-3)/6.
Hint. Substitute n=1,2,3,4,5 directly into the formula and simplify each fraction.
a1=(2-3)/6=-1/6. a2=(4-3)/6=1/6. a3=(6-3)/6=1/2. a4=(8-3)/6=5/6. a5=(10-3)/6=7/6.
✦ Working through each part gives: -1/6, 1/6, 1/2, 5/6, 7/6.
Write the first five terms of the sequence whose nth term is an = (-1)^(n-1).5^(n+1).
Hint. Substitute n=1,2,3,4,5, tracking the alternating sign from (-1)^(n-1) separately from the power of 5.
a1=(-1)^0.5^2=25. a2=(-1)^1.5^3=-125. a3=(-1)^2.5^4=625. a4=(-1)^3.5^5=-3125. a5=(-1)^4.5^6=15625.
✦ Working through each part gives: 25, -125, 625, -3125, 15625.
Write the first five terms of the sequence whose nth term is an = n.(n^2+5)/4.
Hint. Substitute n=1,2,3,4,5 directly into the formula.
a1=1(6)/4=3/2. a2=2(9)/4=9/2. a3=3(14)/4=21/2. a4=4(21)/4=21. a5=5(30)/4=75/2.
✦ Working through each part gives: 3/2, 9/2, 21/2, 21, 75/2.
Find a17 and a24 for the sequence whose nth term is an = 4n - 3.
Hint. Substitute n=17 and n=24 separately.
a17=4(17)-3=68-3=65. a24=4(24)-3=96-3=93.
✦ Working through each part gives: a17=65, a24=93.
Find a7 for the sequence whose nth term is an = n^2/2^n.
Hint. Substitute n=7 directly, computing the numerator and denominator separately before dividing.
Substituting n=7: the numerator is n^2=7^2=49, and the denominator is 2^n=2^7=128. So a7=49/128, which does not simplify further since 49 and 128 share no common factor.
✦ Working through each part gives: 49/128.
Find a9 for the sequence whose nth term is an = (-1)^(n-1).n^3.
Hint. Substitute n=9, checking the sign of (-1)^(n-1) before computing the cube.
Substituting n=9: the exponent n-1=8 is even, so (-1)^8=1, giving a positive sign. The cube term is n^3=9^3=729. So a9=1x729=729.
✦ Working through each part gives: 729.
Find a20 for the sequence whose nth term is an = n(n-2)/(n+3).
Hint. Substitute n=20 directly, computing the numerator and denominator separately before simplifying.
Substituting n=20: the numerator is n(n-2)=20(18)=360, and the denominator is n+3=23. So a20=360/23, which does not simplify further since 23 is prime and does not divide 360.
✦ Working through each part gives: 360/23.
Write the first five terms of the sequence a1=3, an=3a(n-1)+2 for all n>1, and obtain the corresponding series.
Hint. Apply the recurrence repeatedly, using each computed term to find the next.
a1=3. a2=3(3)+2=11. a3=3(11)+2=35. a4=3(35)+2=107. a5=3(107)+2=323.
✦ Working through each part gives: first five terms: 3, 11, 35, 107, 323. Series: 3+11+35+107+323+...
Write the first five terms of the sequence a1=-1, an=a(n-1)/n for n>=2, and obtain the corresponding series.
Hint. Apply the recurrence repeatedly, dividing each previous term by the new index n.
a1=-1. a2=a1/2=-1/2. a3=a2/3=-1/6. a4=a3/4=-1/24. a5=a4/5=-1/120.
✦ Working through each part gives: first five terms: -1, -1/2, -1/6, -1/24, -1/120. Series: -1-1/2-1/6-1/24-1/120+...
Write the first five terms of the sequence a1=a2=2, an=a(n-1)-1 for n>2, and obtain the corresponding series.
Hint. The first two terms are given directly; apply the recurrence from the third term onward.
a1=2, a2=2. a3=a2-1=1. a4=a3-1=0. a5=a4-1=-1.
✦ Working through each part gives: first five terms: 2, 2, 1, 0, -1. Series: 2+2+1+0+(-1)+...
The Fibonacci sequence is defined by a1=a2=1 and an=a(n-1)+a(n-2) for n>2. Find a(n+1)/an for n=1,2,3,4,5.
Hint. First generate enough Fibonacci terms (up to a6), then compute each ratio.
a1=1,a2=1,a3=2,a4=3,a5=5,a6=8. n=1: a2/a1=1/1=1. n=2: a3/a2=2/1=2. n=3: a4/a3=3/2. n=4: a5/a4=5/3. n=5: a6/a5=8/5.
✦ Working through each part gives: 1, 2, 3/2, 5/3, 8/5 for n=1,2,3,4,5 respectively.
Solutions written by the tuition.in editorial team and checked against the NCERT Class 11 Mathematics textbook, Reprint 2026-27 (kemh108.pdf) — Exercise 8.1 (14 questions on general sequences), Exercise 8.2 (32 questions on Geometric Progression — the book's only numbered exercise on G.P.; no 'Exercise 8.3' exists), plus the chapter's Miscellaneous Exercise (18 questions), 64 questions total. Confirmed against the CBSE curriculum PDF that Arithmetic Progression is not re-taught in this chapter (it is Class 10 content; only Arithmetic Mean reappears here) and that the sum to infinity of a G.P. is genuinely CBSE-summative syllabus content despite the current book never deriving it anywhere, including its own Summary — that formula and its recurring-decimal application are taught in the chapter .md but are not sourced from an actual NCERT exercise question, so no corresponding entry appears in this solutions file. Several garbled stacked-fraction and surd expressions (Exercise 8.1 Q4/Q5/Q6/Q8/Q10; Exercise 8.2 Q1/Q5/Q6/Q11) were cross-checked via 300dpi page renders. Miscellaneous Exercise Q16 (simple interest) was deliberately solved as an Arithmetic Progression, not a Geometric one, and flagged as a commonError — a recurring trap since it appears inside a G.P.-focused exercise. Questions are referenced from the NCERT textbook for identification.
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