the sum of first n terms of two APs are in the ratio (3n+8): ( 7n +15), find the ratio of their 12th terms. dont spam please❌❌❌❌❌
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For 1 stAP
Let first term be a common difference be d
sum of n terms s n = 2n(2a+(n−1)d)
nth term a n = a(n−1)d
Similarly for 2 ndAP
Let first term =A common difference be D
Sn = 2n(2A+(n−1)D) & nthterm =An
=A+(n−1)D
We need ratio of 12 th term
i.e., A 12 ofsecondAP
a 12 ofsecond AP
=
A+(12−1)D
a+(12−1)d
=
a+11D
a+11d
It is given that Sumofntermsof 2nd AP
Sumofntermsof 1st 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., 7:1
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