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Answer:
the last three terms =429
\begin{gathered} \begin{aligned} \Rightarrow & & \tt a_{35}+a_{36}+a_{37} & \tt=429 \\ \\ \Rightarrow & & \tt(a+34 d)+(a+35 d) & \tt+(a+36 d) \\ \\ & & &=429 \\ \\ \Rightarrow & & \tt 3 a+105 d & \tt=429 \\ \\ \Rightarrow & & \tt a+35 d & \tt=143 \end{aligned} \end{gathered}
⇒
⇒
⇒
⇒
a
35
+a
36
+a
37
(a+34d)+(a+35d)
3a+105d
a+35d
=429
+(a+36d)
=429
=429
=143
Solving (i) and (ii) , we have
\begin{gathered} \\ \tt a=3 \text { and } d=4 \\ \end{gathered}
a=3 and d=4
\begin{gathered} \\ \\ \begin{array}{l} \therefore \text { Required AP is } \tt a, a+d, a+2 d \text {, } \\ \tt a+3 d, \ldots \\ \tt \text { that is, } 3,3+4,3+2(4), 3+3(4), \ldots \\ \tt \text { or } 3,7,3+8,3+12, \ldots \text { or } 3,7,11, \\ \tt 15, \ldots \end{array} \end{gathered}
∴ Required AP is a,a+d,a+2d,
a+3d,…
that is, 3,3+4,3+2(4),3+3(4),…
or 3,7,3+8,3+12,… or 3,7,11,
15,…