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# In the following AP’s find the missing terms: (i) 2, __ , 26 (ii) __, 13, __, 3 (iii) 5, __, __, $${9}{\dfrac{1}{2}}$$ (iv) –4. __, __, __, __, 6 (v) __, 38, __, __, __, –22

(i) 2, __ , 26

We know that difference between consecutive terms is equal in any A.P.

Let the missing term be x.

$$\Rightarrow$$ x – 2 = 26 – x
$$\Rightarrow$$ 2x = 28
$$\Rightarrow$$ x = 14

Therefore, missing term is 14.

(ii)__, 13, __, 3

Let missing terms be x and y.

The sequence becomes x, 13, y, 3

We know that difference between consecutive terms is constant in any A.P.

$$\Rightarrow$$ y – 13 = 3 – y
$$\Rightarrow$$ 2y = 16
$$\Rightarrow$$ y = 8

And 13 – x = y – 13
$$\Rightarrow$$ x + y = 26
But, we have y = 8,
$$\Rightarrow$$ x + 8 = 26
$$\Rightarrow$$ x = 18

Therefore, missing terms are 18 and 8.

(iii) 5, __, __, $${9}{\dfrac{1}{2}}$$

Here, first term = a = 5 And,
4th term = $$a_4 = {9}{\dfrac{1}{2}}$$

Using formula $$a_n = a + (n - 1)d$$ , to find nth term of arithmetic progression,

$$\Rightarrow a_4$$ = 5 + (4 - 1) d
$$\Rightarrow {{19} \over {2}}$$ = 5 + 3d
$$\Rightarrow$$ 19 = 2 (5 + 3d)
$$\Rightarrow$$ 19 = 10 + 6d
$$\Rightarrow$$ 6d = 19 – 10
$$\Rightarrow$$ 6d = 9
$$\Rightarrow$$ d = $${{3} \over {2}}$$

Therefore, we get common difference
= d = $${{3} \over {2}}$$

Second term
= a + d = $$5 + {{3} \over {2}} = {{13} \over {2}}$$

Third term = second term + d
= $${{13} \over {2}} + {{3} \over {2}} = {{16} \over {2}} = 8$$

Therefore, missing terms are $${{13} \over {2}}$$ and 8

(iv)–4. __, __, __, __, 6

Here, First term = a = –4 and 6th term = $$a_6 = 6$$

Using formula $$a_n = a + (n - 1)d$$ , to find nth term of arithmetic progression,

$$\Rightarrow a_6$$ = -4 + (6 - 1) d
$$\Rightarrow$$ 6 = -4 + 5d
$$\Rightarrow$$ 5d = 10
$$\Rightarrow$$ d = 2

Therefore, common difference = d = 2

Second term = first term + d
= a + d = –4 + 2 = –2
Third term = second term + d
= –2 + 2 = 0
Fourth term = third term + d
= 0 + 2 = 2
Fifth term = fourth term + d
= 2 + 2 = 4

Therefore, missing terms are –2, 0, 2 and 4.

(v) __, 38, __, __, __, –22

We are given 2nd and 6th term.

Using formula $$a_n = a + (n - 1)d$$ , to find nth term of arithmetic progression,

$$a_2$$ = a + (2 - 1) d And $$a_6$$ = a + (6 - 1) d
38 = a + d And -22 = a + 5d

These are equations in two variables, we can solve them using any method.

Using equation (38 = a + d), we can say that a = 38 - d.

Putting value of a in equation (-22 = a + 5d),

$$\Rightarrow$$ -22 = 38 – d + 5d
$$\Rightarrow$$ 4d = -60
$$\Rightarrow$$ d = -15

Using this value of d and putting this in equation 38 = a + d,

38 = a – 15$$\Rightarrow$$ a = 53

Therefore, we get a = 53 and d = -15

First term = a = 53
Third term = second term + d
= 38 – 15 = 23
Fourth term = third term + d
= 23 – 15 = 8
Fifth term = fourth term + d
= 8 – 15 = –7

Therefore, missing terms are 53, 23, 8 and –7.