If the ratio of the sum of the first n term of two A.Ps is (7n + 1) : (4n + 27), then find
the ratio of their 13th terms.
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Step-by-step explanation:
Let the two series' be T
n
and T
n
′
with first terms a and a
′
and common differences d and d
′
The ratio of the sums of the series' S
n
and S
n
′
is given as,
S
n
′
S
n
=
[n/2][2a
′
+[n−1]d
′
]
[n/2][2a+[n−1]d]
=
4n+27
7n+1
Or,
a
′
+[(n−1)/2]d
′
a+[(n−1)/2]d
=
4n+27
7n+1
...(1)
We have to find,
T
11
′
T
11
=
a
′
+10d
′
a+10d
Choosing (n−1)/2=10 or n=21 in (1) we get
T
11
′
T
11
=
a
′
+10d
′
a+10d
=
4(21)+27
7(21)+1
=
111
148
=
3
4
.
Hence, option A.
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