Fill in the blanks in the following table, given that a is the first term, d the common difference and an the nth term of the AP: a d n an (i) 7 3 8 . . . (ii) – 18 . . . 10 0 (iii) . . . – 3 18 – 5 (iv) – 18.9 2.5 . . . 3.6 (v) 3.5 0 105 . . . 62
Step-by-Step Solution
Key Concept: For an arithmetic progression (AP) the nth term is given by the formula \(a_n = a + (n-1)d\). By substituting the known quantities and solving for the missing one, each blank can be filled.
### Row (i)
- Given: \(a = 7\), \(d = 3\), \(n = 8\).
- Use \(a_n = a + (n-1)d\):
\[a_8 = 7 + (8-1)\times 3 = 7 + 7\times 3 = 7 + 21 = 28.\]
- Hence \(a_n = 28\).
### Row (ii)
- Given: \(d = -18\), \(n = 10\), \(a_n = 0\). (The first entry \(a\) is missing.)
- Apply the formula:
\[0 = a + (10-1)(-18) = a - 9\times 18 = a - 162.\]
- Solve for \(a\): \(a = 162\).
### Row (iii)
- Given: \(d = -3\), \(n = 18\), \(a_n = -5\). (The first entry \(a\) is missing.)
- Use the formula:
\[-5 = a + (18-1)(-3) = a - 51.\]
- Hence \(a = 46\).
### Row (iv)
- Given: \(a = -18.9\), \(d = 2.5\), \(a_n = 3.6\). (The term \(n\) is missing.)
- Substitute:
\[3.6 = -18.9 + (n-1)\times 2.5\]
\[ (n-1)\times 2.5 = 3.6 + 18.9 = 22.5 \]
\[ n-1 = \frac{22.5}{2.5} = 9 \]
\[ n = 10.\]
- Therefore \(n = 10\).
### Row (v)
- Given: \(a = 3.5\), \(n = 105\), \(a_n = 62\). (The common difference \(d\) is missing.)
- Apply the formula:
\[62 = 3.5 + (105-1)d = 3.5 + 104d.\]
\[104d = 62 - 3.5 = 58.5\]
\[d = \frac{58.5}{104} = 0.5625 = \frac{9}{16}.\]
- Hence \(d = 0.5625\) (or \(\frac{9}{16}\)).
Filled table
| a | d | n | a_n |
|---|---|---|-----|
| 7 | 3 | 8 | 28 |
| 162 | -18 | 10 | 0 |
| 46 | -3 | 18 | -5 |
| -18.9 | 2.5 | 10 | 3.6 |
| 3.5 | 0.5625 | 105 | 62 |
Correct Answer: {"row_i":{"a":7,"d":3,"n":8,"an":28},"row_ii":{"a":162,"d":-18,"n":10,"an":0},"row_iii":{"a":46,"d":-3,"n":18,"an":-5},"row_iv":{"a":-18.9,"d":2.5,"n":10,"an":3.6},"row_v":{"a":3.5,"d":0.5625,"n":105,"an":62}}