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NDA Math Sequences and Series
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© examsnet.com
Question : 11
Total: 79
If
p
=
(
1111
.
.
.
up to
n
digits
)
, then what is the value of
9
p
2
+
p
?
[NDA-II 2021]
10
n
p
2
p
⋅
10
n
10
n
p
−
1
10
n
p
+
1
Validate
Solution:
👈: Video Solution
(a) Given that,
p
=
(
1111
.
.
.
upto
n
digits
)
=
1
+
10
+
10
2
+
.
.
.
+
10
n
−
1
=
1
(
10
n
−
1
)
10
−
1
[
∵
a
+
a
r
+
a
r
2
+
.
.
.
+
a
r
n
−
1
=
a
(
r
n
−
1
)
r
−
1
]
⇒
p
=
10
n
−
1
9
⇒
9
p
=
10
n
−
1
⇒
9
p
+
1
=
10
n
⇒
9
p
2
+
p
=
10
n
⋅
p
∴
Hence, option (a) is correct.
© examsnet.com
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