Basic Exercises

Solution:

From the general term formula of an arithmetic sequence: $a_3 = a_1 + 2d = 7$ $a_7 = a_1 + 6d = 15$

Subtracting the two equations: $4d = 8$, so $d = 2$

Substituting into the first equation: $a_1 = 7 - 4 = 3$

Answer: $a_1 = 3$, $d = 2$

Solution:

Property of the sum of the first $n$ terms of an arithmetic sequence: $S_5, S_{10} - S_5, S_{15} - S_{10}$ form an arithmetic sequence.

$S_5 = 25$, $S_{10} - S_5 = 75$

The common difference is $75 - 25 = 50$

$S_{15} - S_{10} = 75 + 50 = 125$

$S_{15} = 100 + 125 = 225$

Answer: $S_{15} = 225$

Solution:

By the property of arithmetic sequences: $a_1 + a_9 = 2a_5$

$a_1 + a_5 + a_9 = 2a_5 + a_5 = 3a_5 = 27$

$a_5 = 9$

Answer: $a_5 = 9$

Solution:

$\frac{a_5}{a_2} = q^3 = \frac{48}{6} = 8$

$q = 2$

$a_1 = \frac{a_2}{q} = \frac{6}{2} = 3$

Answer: $q = 2$, $a_1 = 3$

Solution:

When $n = 1$, $a_1 = S_1 = 1$

When $n \geq 2$, $a_n = S_n - S_{n-1} = (2^n - 1) - (2^{n-1} - 1) = 2^{n-1}$

Verify $n = 1$: $a_1 = 2^0 = 1$ ✓

Answer: $a_n = 2^{n-1}$

Solution:

This is an arithmetic sequence with first term $a_1 = 1$, common difference $d = 2$, and last term $a_n = 99$.

Number of terms: $n = \frac{99 - 1}{2} + 1 = 50$

$S_{50} = \frac{50 \times (1 + 99)}{2} = 2500$

Answer: 2500

Solution:

This is a geometric sequence with first term $a_1 = 1$ and common ratio $q = 2$.

$512 = 2^9$, so there are 10 terms.

$S_{10} = \frac{1 \times (2^{10} - 1)}{2 - 1} = 1023$

Answer: 1023

Solution:

Splitting terms: $\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}$

$S = \left(1 - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \cdots + \left(\frac{1}{99} - \frac{1}{100}\right) = 1 - \frac{1}{100} = \frac{99}{100}$

Answer: $\frac{99}{100}$

Solution:

Accumulation method: $a_n - a_1 = \sum_{k=1}^{n-1} 2k = 2 \times \frac{(n-1)n}{2} = n(n-1)$

$a_n = 1 + n^2 - n = n^2 - n + 1$

Answer: $a_n = n^2 - n + 1$

Solution:

This is a geometric sequence with $q = 2$

$a_n = 2 \times 2^{n-1} = 2^n$

Answer: $a_n = 2^n$