For these questions, we need to identify that the series in each case are arithmetic progressions (APs). The formula for the sum of the first $n$ terms of an arithmetic series is:

$S_n = \frac{n}{2} [2a + (n-1) d]$

where:

• $S_n$ is the sum of the first $n$ terms,
• $a$ is the first term of the series,
• $d$ is the common difference between terms,
• $n$ is the number of terms.

Let’s now break down each of the given series.

Question 6:

The series is $1 + 2 + 3 + \cdots = 210$.

• Here, $a = 1$ (the first term).
• The common difference $d = 1$ (since each term increases by 1).
• The sum is $S_n = 210$.

We need to find $n$. Using the formula: $S_n = \frac{n}{2} [2a + (n-1) d] = 210$ $\frac{n}{2} [2(1) + (n-1)(1)] = 210$ $\frac{n}{2} [2 + n - 1] = 210$ $\frac{n}{2} (n+1) = 210$ $n(n+1) = 420$ Solving this quadratic equation: $n^2 + n - 420 = 0$ Using the quadratic formula: $n = \frac{-1 \pm \sqrt{1^2 - 4(1)(-420)}}{2(1)} = \frac{-1 \pm \sqrt{1 + 1680}}{2} = \frac{-1 \pm \sqrt{1681}}{2}$ $n = \frac{-1 \pm 41}{2}$ Thus, $n = 20$ (since $n$ must be positive).

So, 20 terms must be added to get the sum of 210.

Question 7:

The series is $2 + 4 + 6 + \cdots = 420$.

• Here, $a = 2$,
• The common difference $d = 2$,
• The sum $S_n = 420$.

Using the sum formula: $S_n = \frac{n}{2} [2a + (n-1)d] = 420$ $\frac{n}{2} [2(2) + (n-1)(2)] = 420$ $\frac{n}{2} [4 + 2(n-1)] = 420$ $\frac{n}{2} (2n + 2) = 420$ $n(n + 1) = 420$ This gives the same quadratic equation as in Question 6: $n^2 + n - 420 = 0$ So, $n = 20$.

Thus, 20 terms must be added to get the sum of 420.

Question 8:

The series is $90 + 87 + 84 + \cdots = 420$.

• Here, $a = 90$,
• The common difference $d = -3$ (since each term decreases by 3),
• The sum $S_n = 420$.

Using the sum formula: $S_n = \frac{n}{2} [2a + (n-1)d] = 420$ $\frac{n}{2} [2(90) + (n-1)(-3)] = 420$ $\frac{n}{2} [180 - 3(n-1)] = 420$ $\frac{n}{2} [180 - 3n + 3] = 420$ $\frac{n}{2} (183 - 3n) = 420$ Multiplying through: $n(183 - 3n) = 840$ $183n - 3n^2 = 840$ Rearranging: $3n^2 - 183n + 840 = 0$ Dividing through by 3: $n^2 - 61n + 280 = 0$ Solving using the quadratic formula: $n = \frac{-(-61) \pm \sqrt{(-61)^2 - 4(1)(280)}}{2(1)} = \frac{61 \pm \sqrt{3721 - 1120}}{2} = \frac{61 \pm \sqrt{2601}}{2}$ $n = \frac{61 \pm 51}{2}$ Thus, $n = 56 \, \text{or} \, 5$.

Since $n = 5$ makes sense (it's a smaller number of terms), 5 terms must be added to get the sum of 420.

Do you need further details or have any questions?

Here are five related questions to consider:

1. How would the formula for an arithmetic series change if the common difference were negative in every case?
2. Can you derive the sum formula $S_n$ from basic principles of arithmetic sequences?
3. How would you approach finding the number of terms if the sum was instead given as an unknown variable?
4. What are the key differences between arithmetic and geometric progressions in terms of sum calculation?
5. How does the quadratic formula assist in solving sum-related problems in arithmetic progressions?

Tip: When solving quadratic equations arising from sum formulas, remember that only positive values for $n$ are valid since $n$ represents a count of terms.