In the following AP's find the missing terms:
(i)-4. __, __, __, __, 6
(ii) __, 38, __, __, __, -22
(i)
-4. __, __, __, __, 6
Here, First term = a = -4
6th term = a6 = 6
Using formula an=a+ (n−1) d, to find nth term of arithmetic progression, we get
a6=−4+(6−1)d
⇒6=−4+5d
⇒5d=10
⇒ d= 105 =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.
(ii)
__, 38, __, __, __, -22
In this problem, we are given 2nd and 6th term.
Using formula an=a + (n−1) d, to find nth term of arithmetic progression, we get
a2=a+(2−1)d and a6=a+(6−1)d
⇒38=a+d and −22=a+5d
These are equations in two variables; we can solve them using any method. Let’s solve them using substitution method.
Using equation (38=a+ d), we can say that a=38−d. Putting value of a in equation (−22=a+5d), we get
−22=38−d+5d
⇒4d=−60
⇒ d=− 604=−15
Using this value of d and putting this in equation 38=a +d, we get
38=a−15
⇒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.