Replace the words in italics with Noun Clauses :
1. I know him to be honest .
2. She knows the truth.
3. He ambition is to become a doctor.
4. His death was expected.
5. The materiology department predicts a change in weather.
6. The newspaper described the attack by the terrorists.
7. I heard of your success.
8. The judge held the man guilty.
9. He confessed his crime.
10. The meeting place of the kidnappers was known to the police.
Answers
Required AnswErs:-
1. I know him to be honest.
- I know that he is honest.
2. She knows the truth.
- She knows what the truth is.
3. He ambition is to become a doctor.
- His ambition is that he would become a doctor.
4. His death was expected.
- It was expected that he would die.
5. The meteorology department predicts a change in weather.
- The meteorology department predicts that a change would occur in weather.
6. The newspaper described the attack by the terrorists.
- The newspaper described that the attack was caused by the terrorists.
7. I heard of your success.
- I heard that you have succeeded.
8. The judge held the man guilty.
- The judge held that the man was guilt.
9. He confessed his crime.
- He confessed that he made crime.
10. The meeting place of the kidnappers was known to the police.
- Police knew that kidnappers planned of the meeting place.
Answer:
AnswEr:-
The ratio is 53 : 153
GivEn:-
The sum of nth term of two arithmetic series are in ratio of 2n + 3 : 6n + 3
To Find:-
The ratio of their 13th term
SolutiOn:-
Let the sum of one A.P. be Sn and and another one be S'n.
According to the given question,
\implies \sf \: \dfrac{S_{n}}{S'_{n}} = \dfrac{2n + 3}{6n + 3}⟹
S
n
′
S
n
=
6n+3
2n+3
\implies \sf \: \dfrac{ \dfrac{n}{2}[ 2a_{1} + (n - 1)d_{1}]}{ \dfrac{n}{2} [2a_{2} + (n - 1)d_{2}]} = \dfrac{2n + 3}{6n + 3}⟹
2
n
[2a
2
+(n−1)d
2
]
2
n
[2a
1
+(n−1)d
1
]
=
6n+3
2n+3
\implies \sf \: \dfrac{a_{1} + ( \frac{n - 1}{2})d_{1} }{a_{2} + ( \frac{n - 1}{2})d_{2} } = \dfrac{2n + 3}{6n + 3}⟹
a
2
+(
2
n−1
)d
2
a
1
+(
2
n−1
)d
1
=
6n+3
2n+3
Now, we have to find the 13th term of each A.P.
Therefore,
\sf \: (\dfrac{n - 1}{2} ) = 12(
2
n−1
)=12
\implies \sf \: n - 1 = 12 \times 2⟹n−1=12×2
\implies \sf \: n = 24 + 1⟹n=24+1
\boxed { \red{\sf{ \implies \: n = 25}}}
⟹n=25
Now, putting the value of n we get,
\implies \sf \: \dfrac{a_{1} + 12d_{1} }{a_{2} + 12d_{2} } = \dfrac{(2 \times 25)+ 3}{(6 \times 25) + 3}⟹
a
2
+12d
2
a
1
+12d
1
=
(6×25)+3
(2×25)+3
\implies \sf \: \dfrac{a_{1} + 12d_{1} }{a_{2} + 12d_{2} } = \dfrac{50+ 3}{150 + 3}⟹
a
2
+12d
2
a
1
+12d
1
=
150+3
50+3
\boxed{ \red{ \sf{ \implies \: \frac{t_{n}}{t'_{n}} = \dfrac{53}{153} }}}
⟹
t
n
′
t
n
=
153
53
Therefore, the ratio is 53 : 153