Arithmetic Progressions — Maths STD 10 — Question

Let the first terms and the common difference of the given AP be a and d respectively.
$ \Rightarrow $ a + (2 - 1)d = 38 $\because {a_n} = a + (n - 1)d$
$ \Rightarrow $ a + d = 38 ....... (1)
Sixth term = -22
$ \Rightarrow $ a + (6 - 1) d = -22
$ \Rightarrow $ a + 5d = -22 ......... (2)
Solving (1) and (2), we get
a = 53
d = -15
Therefore,
Third term = 53 + (3 - 1) (-15) $\because {a_n} = a + (n - 1)d$
= 53 - 30
= 23
Fourth term = 53 + (4 - 1) (-15) $\because {a_n} = a + (n - 1)d$
= 8
Fifth = 53 + (5 - 1) (-15) $\because {a_n} = a + (n - 1)d$
= -7
Hence, the missing terms in the boxes are 53, 23, 8, -7