The ratio between 5th term and 8th term of an arithmetic sequence is 5:11. Then find the ratio between 6th term and 9th term?
Answers
Let
′
a
′
and
′
d
′
arefirsttermand
Common \: difference \: of \: an \: A.P .CommondifferenceofanA.P.
/* We know that , */
\boxed{\pink{ n^{th} \: term (a_{n}) = a + (n-1)d }}
n
th
term(a
n
)=a+(n−1)d
Ratio \: of \: between \: 5^{th} \: term \: andRatioofbetween5
th
termand
8^{th} \: term = 5 : 11 \: ( given )8
th
term=5:11(given)
\implies \frac{a_{5}}{a_{8}} = \frac{5}{11}⟹
a
8
a
5
=
11
5
\implies \frac{a + 4d }{a + 7d } = \frac{5}{11}⟹
a+7d
a+4d
=
11
5
\implies 11(a+4d) = 5(a+7d)⟹11(a+4d)=5(a+7d)
\implies 11a + 44d = 5a + 35d⟹11a+44d=5a+35d
\implies 11a - 5a = 35d - 44d⟹11a−5a=35d−44d
\implies 6a = -9d⟹6a=−9d
\implies \frac{a}{d} = \frac{-9 }{6}⟹
d
a
=
6
−9
\implies \frac{a}{d} = \frac{-3}{2} \: --(1)⟹
d
a
=
2
−3
−−(1)
Now, The \: ratio \: between \: 6^{th} \:termNow,Theratiobetween6
th
term
and \: 9^{th} \: term = \frac{a_{6}}{a_{9}}and9
th
term=
a
9
a
6
= \frac{a + 5d }{a + 8d}=
a+8d
a+5d
/* Dividing numerator and denominator by d, we get */
= \frac{\frac{a}{d} + 5 }{\frac{a}{d} + 8}=
d
a
+8
d
a
+5
= \frac{ \frac{-3}{2} + 5}{\frac{-3}{2} + 8 }=
2
−3
+8
2
−3
+5
= \frac{ \frac{-3+10}{2}}{\frac{-3+16}{2}}=
2
−3+16
2
−3+10
= \frac{ 7}{13}=
13
7
= 7 : 13=7:13
Therefore.,
\red{ The \: ratio \: between \: 6^{th} \:term}Theratiobetween6
th
term
\red{and \: 9^{th} \: term}\green { = 7: 13}and9
th
term=7:13
Answer:
Let a ′ and d
′
arefirsttermand
Common \: difference \: of \: an \: A.P .CommondifferenceofanA.P.
/* We know that , */
\boxed{\pink{ n^{th} \: term (a_{n}) = a + (n-1)d }}
n
th
term(a
n
)=a+(n−1)d
Ratio \: of \: between \: 5^{th} \: term \: andRatioofbetween5
th
termand
8^{th} \: term = 5 : 11 \: ( given )8
th
term=5:11(given)
\implies \frac{a_{5}}{a_{8}} = \frac{5}{11}⟹
a 8
a 5
= 11 5
\implies \frac{a + 4d }{a + 7d } = \frac{5}{11}⟹
a+7d
a+4d
=
11
5
\implies 11(a+4d) = 5(a+7d)⟹11(a+4d)=5(a+7d)
\implies 11a + 44d = 5a + 35d⟹11a+44d=5a+35d
\implies 11a - 5a = 35d - 44d⟹11a−5a=35d−44d
\implies 6a = -9d⟹6a=−9d
\implies \frac{a}{d} = \frac{-9 }{6}⟹
d
a
=
6
−9
\implies \frac{a}{d} = \frac{-3}{2} \: --(1)⟹
d
a
=
2
−3
−−(1)
Now, The \: ratio \: between \: 6^{th} \:termNow,Theratiobetween6
th
term
and \: 9^{th} \: term = \frac{a_{6}}{a_{9}}and9
th
term=
a
9
a
6
= \frac{a + 5d }{a + 8d}=
a+8d
a+5d
/* Dividing numerator and denominator by d, we get */
= \frac{\frac{a}{d} + 5 }{\frac{a}{d} + 8}=
d
a
+8
d
a
+5
= \frac{ \frac{-3}{2} + 5}{\frac{-3}{2} + 8 }=
2
−3
+8
2
−3
+5
= \frac{ \frac{-3+10}{2}}{\frac{-3+16}{2}}=
2
−3+16
2
−3+10
= \frac{ 7}{13}=
13
7
= 7 : 13=7:13
Therefore.,
\red{ The \: ratio \: between \: 6^{th} \:term}Theratiobetween6
th
term
\red{and \: 9^{th} \: term}\green { = 7: 13}and9
th
term=7:13