the ratio of the sum use of n terms of two ap is 5 and + 4 is 29 and + 6 find the ratio of their 25th comes
Answers
Answer:
Let us assume that for the first AP, the first term is a and common difference is d
For the second AP, the first term is A and the common difference is D,
Now as per the problem,
\frac{\frac{n}{2}[2 a+(n-1) d]}{\frac{n}{2}[2 A+(n-1) D]}=\frac{5 n+4}{9 n+6}
2
n
[2A+(n−1)D]
2
n
[2a+(n−1)d]
=
9n+6
5n+4
\frac{[2 a+(n-1) d]}{[2 A+(n-1) D]}=\frac{5 n+4}{9 n+6}
[2A+(n−1)D]
[2a+(n−1)d]
=
9n+6
5n+4
\frac{\left[a+\frac{(n-1) d}{2}\right]}{\left[A+\frac{(n-1) D}{2}\right]}=\frac{5 n+4}{9 n+6}
[A+
2
(n−1)D
]
[a+
2
(n−1)d
]
=
9n+6
5n+4
……………….. (i)
Now the ratio of 18th term of both the ap is =\frac{a+17 d}{A+17 D}=
A+17D
a+17d
….. (ii)
Hence \frac{n-1}{2}=17
2
n−1
=17
n=35
Now putting value of n in equation (i), we get
\frac{\left[a+\frac{(35-1) d}{2}\right]}{\left[A+\frac{(35-1) D}{2}\right]}=\frac{5 * 35+4}{9 * 35+6}
[A+
2
(35−1)D
]
[a+
2
(35−1)d
]
=
9∗35+6
5∗35+4
\frac{a+17 d}{A+17 D}=\frac{179}{321}
A+17D
a+17d
=
321
179
So the ratio of 18^{\text {th }}18
the ans is 179:321