The image contains mathematical problems primarily about arithmetic progressions (AP) in the Nepali language. Here's the structured analysis and solutions to the problems presented in the image:

Problem 4: Identifying AP sequences and finding $x$:

You are tasked with determining which sequences are arithmetic progressions (AP).

(a) $5, x, 9, \ldots$

• The difference between consecutive terms in an AP is constant.
• $x - 5 = 9 - x \implies 2x = 14 \implies x = 7$.
• Sequence: $5, 7, 9, \ldots$ (This is an AP).

(b) $x + 1, x + 5, 3x + 1, \ldots$

• Differences: $(x + 5) - (x + 1) = 4$ and $(3x + 1) - (x + 5) = 2x - 4$.
• For it to be an AP: $4 = 2x - 4 \implies 2x = 8 \implies x = 4$.
• Sequence: $5, 9, 13, \ldots$ (This is an AP when $x = 4$).

(c) $x + 2, 3x, 4x + 1, \ldots$

• Differences: $3x - (x + 2) = 2x - 2$ and $(4x + 1) - 3x = x + 1$.
• For it to be an AP: $2x - 2 = x + 1 \implies x = 3$.
• Sequence: $5, 9, 13, \ldots$ (This is an AP when $x = 3$).

Problem 5: Finding sums of AP terms

(a) Sum of the first 20 terms: $7 + 11 + 15 + 19 + \ldots$

• First term ($a$) = 7, Common difference ($d$) = 4, $n = 20$.
• Sum formula: $S_n = \frac{n}{2} [2a + (n - 1)d]$.
• $S_{20} = \frac{20}{2} [2(7) + (20 - 1)4]$ $S_{20} = 10 [14 + 76] = 10 \cdot 90 = 900.$

(b) Sum of 7 terms: $4 - 1 - 6 - 11 - 16 - \ldots$

• $a = 4$, $d = -5$, $n = 7$.
• $S_7 = \frac{7}{2} [2(4) + (7 - 1)(-5)]$. $S_7 = \frac{7}{2} [8 - 30] = \frac{7}{2} \cdot (-22) = -77.$

(c) Sum of first 13 terms: $5 + 10 + 15 + \ldots + 65$

• $a = 5$, $d = 5$, Last term ($l$) = 65.
• $n = \frac{l - a}{d} + 1 = \frac{65 - 5}{5} + 1 = 13$.
• Sum formula: $S_n = \frac{n}{2} (a + l)$. $S_{13} = \frac{13}{2} (5 + 65) = \frac{13}{2} \cdot 70 = 455.$

(d) Sum of 16 terms: $-64, -48, -32, \ldots, 32$

• $a = -64$, $l = 32$, $d = 16$.
• $n = \frac{l - a}{d} + 1 = \frac{32 - (-64)}{16} + 1 = 7 + 1 = 8$.
• $S_8 = \frac{8}{2} (-64 + 32) = 4 \cdot (-32) = -128.$

(e) Sum of first 10 odd numbers

• Odd numbers: $1, 3, 5, \ldots$.
• $a = 1$, $d = 2$, $n = 10$.
• $S_{10} = \frac{10}{2} [2(1) + (10 - 1)(2)] = 5 [2 + 18] = 5 \cdot 20 = 100.$

(f) Sum of first 100 natural numbers:

• Natural numbers: $1, 2, 3, \ldots$.
• $S_{100} = \frac{100}{2} (1 + 100) = 50 \cdot 101 = 5050.$

(g) Sum of natural numbers from 50 to 100:

• Numbers: $50, 51, 52, \ldots, 100$.
• $a = 50$, $l = 100$, $n = 100 - 50 + 1 = 51$.
• $S_{51} = \frac{51}{2} (50 + 100) = \frac{51}{2} \cdot 150 = 3825.$

Problem at the Bottom

Find first term ($a$), last term ($l$), and number of terms ($n$):

• First term = $a = 1$, Last term = $l = 50$, Number of terms = $n = 50$.
• Sum formula: $S_{50} = \frac{50}{2} (1 + 50) = 25 \cdot 51 = 1275.$

Follow-up Questions:

1. How do you find the $n$-th term of an arithmetic progression?
2. Can you explain why the sum of the first $n$ odd numbers is always $n^2$?
3. How can we confirm if a given sequence is an AP?
4. What happens to the sum formula if the common difference ($d$) is negative?
5. Can you derive the formula for the sum of an AP directly from the sequence?

Tip: Always double-check your common difference ($d$) and the number of terms ($n$) when calculating sums in an AP.