An AP starts with a positive fraction and every alternate term is an integer. If the sum of the fi rst 11 terms is 33, then the
fourth term is [1]
(a) 2 (b) 3
(c) 5 (d) 6
Answers
Answered by
0
Answer:
Step-by-step explanation:
Then, Sum to
n
terms of an AP
=
n
2
(
2
a
+
(
n
−
1
)
d
)
Given,
S
11
=
33
=>
n
2
(
2
a
+
(
n
−
1
)
d
)
=
33
=>
11
2
(
2
a
+
(
11
−
1
)
d
)
=
33
Solving we get
a
+
5
d
=
3
As
a
is a fraction,
d
should be the same fraction, so that adding
a
+
d
gives an integer second term.
So,
a
+
5
a
=
3
=>
a
=
1
2
=
d
n
t
h
term of the AP
T
n
=
a
+
(
n
−
1
)
d
So,
T
4
=
1
2
+
(
4
−
1
)
1
2
=
2
Answered by
3
Step-by-step explanation:
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