let sum of first n term of two ap are in the ratio 2n+1/3n-1 . Find ratio of there 11 th term .....plz someone tell na
Share
let sum of first n term of two ap are in the ratio 2n+1/3n-1 . Find ratio of there 11 th term .....plz someone tell na
Sign Up to our social questions and Answers Engine to ask questions, answer people’s questions, and connect with other people.
Login to our social questions & Answers Engine to ask questions answer people’s questions & connect with other people.
[tex] \sf \to \green{{ \underline{ \underline{answer \ratio}}}} \\ \\ \sf \to \: 11th \: term \: of \: an \: ap \: = \frac{43}{62} [/tex]
[tex] \sf \to \: \orange{{ \underline { \underline{step - \: by - \: step - \: explanation \ratio}}}}[/tex]
[tex] \sf \to \: sum \: of \: ratio \: of \: n \: terms \: = \dfrac{2n + 1}{3n - 1} \\ \\ \sf \to \: formula \: of \: sum \: of \: n \: terms \\ \\ \sf \to \: s_{n} \: = \dfrac{n}{2} (2a \: + (n - 1)d) \\ \\ \sf \to \: \frac{ s_{n} }{ s_{n} {}^{1} } = \frac{ \dfrac{n}{2}(2a \: + (n - 1)d }{ \frac{n}{2}(2a {}^{1} + (n - 1)d {}^{1} } = \frac{2n + 1}{3n - 1} \\ \\ \sf \to \: \frac{ s_{n} }{ s_{n} {}^{1} } = \frac{2a + (n - 1)d}{2a {}^{1} + (n - 1) {d}^{1} } \\ \\ \sf \to \: \frac{ s_{n} }{ s_{n} {}^{1} } = \frac{a \: + ( \dfrac{n - 1}{2})d }{ {a}^{1} + (\dfrac{n - 1}{2} )d {}^{1} } \\ \\ \sf \to \: 11th \: term \: of \: an \: ap \: = a \: + 10d \\ \\ \sf \to \: \dfrac{ t_{11 } }{ t_{11} {}^{1} } = \dfrac{a + 10d}{ {a}^{1} + 10d {}^{1} } \\ \\ \sf \to \: \frac{n - 1}{2} = 10 \\ \\ \sf \to \: n \: = 21 \\ \\ \sf \to \: \dfrac{ t_{11} }{ t_{11 {}^{1} }} = \dfrac{a + 10d}{ {a}^{1} + 10d {}^{1} } = \: \dfrac{2n + 1}{3n - 1} \\ \\ \sf \to \: \frac{ t_{11}}{ t_{11} {}^{1} } = \frac{2 \times 21 + 1}{3 \times 21 - 1 } = \frac{43}{62} \\ \\ \sf \to \: \green{{ \underline{11th \: term \: of \: an \: ap \: in \: ratio \: = \frac{43}{62} }}}[/tex]
Answer:
Step-by-step explanation:
Let the two series' be T
n
and T
n
′
with first terms a and a
′
and common differences d and d
′
The ratio of the sums of the series' S
n
and S
n
′
is given as,
S
n
′
S
n
=
[n/2][2a
′
+[n−1]d
′
]
[n/2][2a+[n−1]d]
=
4n+27
7n+1
Or,
a
′
+[(n−1)/2]d
′
a+[(n−1)/2]d
=
4n+27
7n+1
...(1)
We have to find,
T
11
′
T
11
=
a
′
+10d
′
a+10d
Choosing (n−1)/2=10 or n=21 in (1) we getT
11
=
a
′
+10d
′
a+10d
=
4(21)+27
7(21)+1
=
111
148
=
3
4
.
Hence, option A.