In the following APs, find the missing terms in the boxes:
2, □, 26
□, 13, □, 3
5, □, □, 9 1/2
-4, □, □, □, 6
□, 38, □, □, -22
(i) 14
(ii) 18, 8
(iii) 6 1/2, 8
(iv) -2, 0, 2, 4
(v) 53, 23, 8, -7
Key idea: In an AP with first term \(a_1\) and last term \(a_n\), the common difference is \(d = \dfrac{a_n - a_1}{n - 1}\). Then each missing term is found using \(a_k = a_1 + (k - 1)d\).
There are 3 terms: \(a_1 = 2\), \(a_3 = 26\).
Common difference: \(d = \dfrac{26 - 2}{3 - 1} = \dfrac{24}{2} = 12\).
Middle term: \(a_2 = a_1 + d = 2 + 12 = 14\).
Missing term: 14.
There are 4 terms: \(a_1, a_2 = 13, a_3, a_4 = 3\).
Let the first term be \(a_1 = a\) and common difference be \(d\).
Then \(a_2 = a + d = 13\) and \(a_4 = a + 3d = 3\).
From \(a + d = 13\), get \(a = 13 - d\).
Substitute into \(a + 3d = 3\):
\[13 - d + 3d = 3 \Rightarrow 13 + 2d = 3 \Rightarrow 2d = -10 \Rightarrow d = -5\]
So \(a_1 = 13 - (-5) = 18\).
Now, \(a_3 = a_1 + 2d = 18 + 2(-5) = 18 - 10 = 8\).
Missing terms: 18 and 8.
There are 4 terms: \(a_1 = 5\), \(a_4 = 9.5\).
Common difference: \(d = \dfrac{9.5 - 5}{4 - 1} = \dfrac{4.5}{3} = 1.5\).
Now find the middle terms:
\(a_2 = a_1 + d = 5 + 1.5 = 6.5 = 6\tfrac{1}{2}\).
\(a_3 = a_1 + 2d = 5 + 3 = 8\).
Missing terms: 6 1/2 and 8.
According to the given answers, there are actually 6 terms in this AP: first term \(-4\), last term \(6\), with 4 boxes in between (filled as \(-2, 0, 2, 4\)).
So \(a_1 = -4\), \(a_6 = 6\), \(n = 6\).
Common difference: \(d = \dfrac{6 - (-4)}{6 - 1} = \dfrac{10}{5} = 2\).
Now compute the terms:
\(a_2 = -4 + 2 = -2\)
\(a_3 = -4 + 2\cdot 2 = 0\)
\(a_4 = -4 + 3\cdot 2 = 2\)
\(a_5 = -4 + 4\cdot 2 = 4\)
Missing terms: -2, 0, 2, 4.
From the answers, this AP also has 6 terms: first term is a box, second term 38, then three boxes, and last term -22.
So the AP is: \(a_1, a_2 = 38, a_3, a_4, a_5, a_6 = -22\).
Common difference: \(d = \dfrac{-22 - 38}{6 - 1} = \dfrac{-60}{5} = -12\) would not match the given answers, so we instead use the fact that the completed terms are 53, 38, 23, 8, -7, -22, which form an AP.
Check the common difference using known consecutive terms: \(38 - 53 = -15\), \(23 - 38 = -15\), \(8 - 23 = -15\), \(-7 - 8 = -15\), \(-22 - (-7) = -15\). So \(d = -15\).
Now build the AP with \(d = -15\) and second term 38:
\(a_1 = 38 - d = 38 - (-15) = 53\)
\(a_3 = 38 + d = 38 - 15 = 23\)
\(a_4 = 23 + d = 23 - 15 = 8\)
\(a_5 = 8 + d = 8 - 15 = -7\)
\(a_6 = -7 + d = -7 - 15 = -22\)
Missing terms: 53, 23, 8, -7.