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Sagot :
Answer:In this problem,
a
n
=x
a
1
=1
d=6−1=5
∴x=1+(n−1)×5
∴x−1=(n−1)×5
∴(n−1)=
5
x−1
∴n=
5
x−1
+1
∴n=
5
x−1+5
∴n=
5
x+4
Equation (1)
2) Sum of n terms of A.P. is
∑a
n
=
2
n
[2a+(n−1)d]
As given in problem, ∑a
n
=148
∴
2
n
[2a+(n−1)d]=148
∴
2
(
5
x+4
)
[(2×1)+(
5
x+4
−1)5]=148
∴
10
x+4
[2+(
5
x+4−5
)5]=148
∴
10
x+4
[2+x−1]=148
∴
10
x+4
[x+1]=148
∴(x+4)(x+1)=1480
∴x
2
+5x+4=1480
∴x
2
+5x−1476=0
∴x=
2×1
−5±
(5)
2
−4×1×(−1476)
∴x=
2
−5±
25+5904
∴x=
2
−5±
5929
∴x=
2
−5±77
neglecting negative root, we get,
x=
2
−5+77
∴x=
2
72
∴x=36
Step-by-step explanation: i hope it was helpfu;;
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