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Math B

Arithmetic and Geometric Sequences: Practice Problems

Arithmetic and geometric sequences are the foundation of the topic. Practise the formulas for ana_n and SnS_n and how to find the first term, difference or ratio from given conditions.

Basic, Standard, Advanced: Grade 11 · Term 2

Math problem generator

Level

Arithmetic sequences

A sequence with a constant difference dd between consecutive terms is arithmetic.

Formulas
an=a1+(n−1)da_n = a_1 + (n-1)d
Sn=12n(a1+an)=12n{2a1+(n−1)d}S_n = \frac{1}{2}n(a_1 + a_n) = \frac{1}{2}n\{2a_1 + (n-1)d\}

Example: with first term 33 and difference 44, an=4n−1a_n = 4n - 1 and S10=12⋅10⋅(3+39)=210S_{10} = \frac{1}{2} \cdot 10 \cdot (3 + 39) = 210.

Arithmetic mean

a,b,ca, b, c form an arithmetic sequence in this order   ⟺  2b=a+c\iff 2b = a + c.

Geometric sequences

A sequence with a constant ratio rr between consecutive terms is geometric.

Formulas
an=a1rn−1a_n = a_1r^{n-1}
Sn=a1(rn−1)r−1=a1(1−rn)1−r(r≠1),Sn=na1(r=1)S_n = \frac{a_1(r^n - 1)}{r - 1} = \frac{a_1(1 - r^n)}{1 - r}\quad (r \neq 1),\qquad S_n = na_1\quad (r = 1)
Geometric mean

a,b,ca, b, c form a geometric sequence in this order   ⟺  b2=ac\iff b^2 = ac (so bb may have two values, ±\pm).

Finding a sequence from conditions

  • Two given terms: let the first term be aa and the difference dd (or ratio rr), and solve the simultaneous equations. For geometric sequences, dividing one equation by the other eliminates aa.
  • Maximum sum: if the first term is positive and the difference negative, the sum is largest once all positive terms are added. Find the nn with an>0a_n > 0.
  • Common terms: the numbers shared by two arithmetic sequences recur every lcm of the two differences.
  • Three numbers: write them as a−d, a, a+da - d,\ a,\ a + d (arithmetic) or ar, a, ar\frac{a}{r},\ a,\ ar (geometric).

Savings plans

Depositing aa at the start of every year with annual interest rate rr (compounded yearly), the total at the end of year nn is

a(1+r)+a(1+r)2+⋯+a(1+r)n=a(1+r){(1+r)n−1}r,a(1 + r) + a(1 + r)^2 + \cdots + a(1 + r)^n = \frac{a(1 + r)\{(1 + r)^n - 1\}}{r},

a geometric sum: earlier deposits earn interest for more years.