The sum of n terms of two arithmetic progressions are in the ratio of (6n + 2) : (9n – 51).
Find the ratio of their 16th terms.
Answers
Answered by
38
Answer:
the ratio is 35 I hope it helpssssss⊙﹏⊙❤️❤️
Step-by-step explanation:
S
n
=
9n+6
5n+4
2
n
[2a
′
+(n−1)d
′
]
2
n
[2a+(n−1)d]
=
9n+6
5n+4
a+(
2
n−1
)d
′
a+(
2
n−1
)d
=
9n+6
5n+4
……..(i)
a+17d
′
a+17d
=
9(35)+6
5(35)+4
=
321
179
∴
T
18
′
T
18
=
321
179
∴
T
18
′
T
18
=
a
′
+17d
′
a+17d
……….(ii)
Comparing equation (i) and (ii),
2
n−1
=17
∴n−1=34
n=35.
Answered by
0
Answer:
S
n
′
S
n
=
9n+6
5n+4
2
n
[2a
′
+(n−1)d
′
]
2
n
[2a+(n−1)d]
=
9n+6
5n+4
a+(
2
n−1
)d
′
a+(
2
n−1
)d
=
9n+6
5n+4
……..(i)
a+17d
′
a+17d
=
9(35)+6
5(35)+4
=
321
179
∴
T
18
′
T
18
=
321
179
∴
T
18
′
T
18
=
a
′
+17d
′
a+17d
……….(ii)
Comparing equation (i) and (ii),
2
n−1
=17
∴n−1=34
n=35.
Attachments:
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