In the following APs find the missing term in the boxes.
(i) 2, ...., 26
(ii) ....., 13, ....., 3
(iii) 5, ....., ....., \(9\frac{1}{2}\)
(iv) -4, ....., ....., ....., ......, 6
(v) ....., 38, ....., ....., ......, -22
(i) For the given A.P., 2,2 , 26
The first and third term are;
a = 2
a3 = 26
As we know, for an A.P.,
an = a+(n −1)d
Therefore, putting the values here,
a3 = 2+(3-1)d
26 = 2+2d
24 = 2d
d = 12
a2 = 2+(2-1)12
= 14
Therefore, 14 is the missing term.
(ii) For the given A.P., , 13, ,3
a2 = 13 and
a4 = 3
As we know, for an A.P.,
an = a+(n−1) d
Therefore, putting the values here,
a2 = a +(2-1)d
13 = a+d ………………. (i)
a4 = a+(4-1)d
3 = a+3d ………….. (ii)
On subtracting equation (i) from (ii), we get,
– 10 = 2d
d = – 5
From equation (i), putting the value of d,we get
13 = a+(-5)
a = 18
a3 = 18+(3-1)(-5)
= 18+2(-5) = 18-10 = 8
Therefore, the missing terms are 18 and 8 respectively.
(iii) For the given A.P.,
a = 5 and
a4 = 19/2
As we know, for an A.P.,
an = a+(n−1)d
Therefore, putting the values here,
a4 = a+(4-1)d
19/2 = 5+3d
(19/2) – 5 = 3d
3d = 9/2
d = 3/2
a2 = a+(2-1)d
a2 = 5+3/2
a2 = 13/2
a3 = a+(3-1)d
a3 = 5+2×3/2
a3 = 8
Therefore, the missing terms are 13/2 and 8 respectively.
(iv) For the given A.P.,
a = −4 and
a6 = 6
As we know, for an A.P.,
an = a +(n−1) d
Therefore, putting the values here,
a6 = a+(6−1)d
6 = − 4+5d
10 = 5d
d = 2
a2 = a+d = − 4+2 = −2
a3 = a+2d = − 4+2(2) = 0
a4 = a+3d = − 4+ 3(2) = 2
a5 = a+4d = − 4+4(2) = 4
Therefore, the missing terms are −2, 0, 2, and 4 respectively.
(v) For the given A.P.,
a2 = 38
a6 = −22
As we know, for an A.P.,
an = a+(n −1)d
Therefore, putting the values here,
a2 = a+(2−1)d
38 = a+d ……………………. (i)
a6 = a+(6−1)d
−22 = a+5d …………………. (ii)
On subtracting equation (i) from (ii), we get
− 22 − 38 = 4d
−60 = 4d
d = −15
a = a2 − d = 38 − (−15) = 53
a3 = a + 2d = 53 + 2 (−15) = 23
a4 = a + 3d = 53 + 3 (−15) = 8
a5 = a + 4d = 53 + 4 (−15) = −7
Therefore, the missing terms are 53, 23, 8, and −7 respectively.