the sum of n terms of two arithmetic progressions are in the ratio (3n+8) : (7n+ 15). find the ratio of their 12 term
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Answer:
For 1
st
AP
Let first term be a common difference be d
sum of n terms s
n
=
2
n
(2a+(n−1)d)
n
th
term a
n
=a(n−1)d
Similarly for 2
nd
AP
Let first term =A common difference be D
S
n
=
2
n
(2A+(n−1)D) & n
th
term =A
n
=A+(n−1)D
We need ratio of 12
th
term
i.e.,
A
12
ofsecondAP
a
12
offirstAP
=
A+(12−1)D
a+(12−1)d
=
a+11D
a+11d
It is given that
Sumofntermsof2
nd
AP
Sumofntermsof1
st
AP
=
7n+15
3n+8
∴
2A+(n−1)D
2a+(n−1)d
=
7n+15
3n+8
………….(1)
2(A+(
2
n−1
)D)
2(a+(
2
n−1
)d)
=
7n+15
3n+8
………………(1)
we need to find
A+11D
a+11d
Hence
2
n−1
=11
n−1=22
n=23
Putting n=23 in (1)
A+(
2
23−1
)D
a+(
2
23−1
)d
=
7×23+15
3×23+18
∴
A+11D
a+11d
=
16
7
Hence ratio of their 12
th
term is
16
7
i.e., 7:16.
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