solve mate plzzz and dont give unnecessiary answer
Answers
Step-by-step explanation:
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.
HEY MATE
HERE IS YOUR
ANSWER
PLSS MARK IT AS BRAINLIST
answer is 7:16
have a great day
