sum of first n terms of 2 APs are in the ratio n+7 : 3n+11 then the ratio of 9th term is
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Step-by-step explanation:
Let a
1
, a
2
be the first terms and d
1
, d
2
the common differences of the two given A.P.'s. Then, the sums of their n terms are given by
S
n
=
2
n
[2a
1
+(n−1)d
1
]
And,
S
n
′
=
2
n
[2a
2
+(n−1)d
2
]
Therefore,
S
n
′
S
n
=
2
n
[2a
2
+(n−1)d
2
]
2
n
[2a
1
+(n−1)d
1
]
=
2a
2
+(n−1)d
2
2a
1
+(n−1)d
1
It is given that
S
n
′
S
n
=
4n+27
7n+1
⟹
2a
2
+(n−1)d
2
2a
1
+(n−1)d
1
=
4n+27
7n+1
....(1)
To find the ratio of the mth terms of the two given AP's, we replace n by (2m-1) in equation 1.
Therefore,
2a
2
+(2m−2)d
2
2a
1
+(2m−2)d
1
=
4(2m−1)+27
7(2m−1)+1
⟹
a
2
+(m−1)d
2
a
1
+(m−1)d
1
=
8m+23
14m−6
Hence, the ratio of the mth terms of two AP's is (14m−6):(8m+23).
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