If the travel time of the pot in front of a 4 meter long window is 0.4 seconds, what is the initial velocity of the pot at the bottom edge of the window, how many meters per second?
Answers
Answer:
Let assume that
First term of an AP series is a
and
Common difference of an AP is d
Given that,
\begin{gathered}\rm \: a_5 = 16 \\ \end{gathered}
a
5
=16
Wᴇ ᴋɴᴏᴡ ᴛʜᴀᴛ,
↝ nᵗʰ term of an arithmetic progression is,
\begin{gathered}\begin{gathered}\red\bigstar\:\:{\underline{\orange{\boxed{\bf{\green{a_n\:=\:a\:+\:(n\:-\:1)\:d}}}}}} \\ \end{gathered}\end{gathered}
★
a
n
=a+(n−1)d
Wʜᴇʀᴇ,
aₙ is the nᵗʰ term.
a is the first term of the progression.
n is the no. of terms.
d is the common difference.
Tʜᴜs,
\begin{gathered}\rm \: a + (5 - 1)d = 16 \\ \end{gathered}
a+(5−1)d=16
\begin{gathered}\rm \: a + 4d = 16 \\ \end{gathered}
a+4d=16
\begin{gathered}\rm\implies \:a = 16 - 4d - - - (1) \\ \end{gathered}
⟹a=16−4d−−−(1)
Also given that,
\begin{gathered}\rm \: a_{13} \: = \: 4a_3 \\ \end{gathered}
a
13
=4a
3
\begin{gathered}\rm \: a + 12d = 4(a + 2d) \\ \end{gathered}
a+12d=4(a+2d)
\begin{gathered}\rm \: a + 12d = 4a + 8d\\ \end{gathered}
a+12d=4a+8d
\begin{gathered}\rm \: 4d = 3a\\ \end{gathered}
4d=3a
On substituting the value of a from equation (1), we get
\begin{gathered}\rm \: 4d = 3(16 - 4d)\\ \end{gathered}
4d=3(16−4d)
\begin{gathered}\rm \: 4d = 48 - 12d\\ \end{gathered}
4d=48−12d
\begin{gathered}\rm \: 16d = 48 \\ \end{gathered}
16d=48
\begin{gathered}\rm\implies \:d \: = \: 3 \\ \end{gathered}
⟹d=3
On substituting value of d in equation (1), we get
\begin{gathered}\rm \: a = 16 - 4 \times 3 \\ \end{gathered}
a=16−4×3
\begin{gathered}\rm \: a = 16 - 12 \\ \end{gathered}
a=16−12
\begin{gathered}\rm\implies \:a = 4 \\ \end{gathered}
⟹a=4
Now,
Wᴇ ᴋɴᴏᴡ ᴛʜᴀᴛ,
↝ Sum of n terms of an arithmetic progression is,
\begin{gathered}\begin{gathered}\red\bigstar\:\:{\underline{\orange{\boxed{\bf{\green{S_n\:=\dfrac{n}{2} \bigg(2 \:a\:+\:(n\:-\:1)\:d \bigg)}}}}}} \\ \end{gathered}\end{gathered}
★
S
n
=
2
n
(2a+(n−1)d)
Wʜᴇʀᴇ,
Sₙ is the sum of n terms of AP.
a is the first term of the progression.
n is the no. of terms.
d is the common difference.
Now, we have
First term of an AP, a = 4
Common difference of an AP, d = 3
Number of terms, n = 10
So,
\rm \: S_{10} = \dfrac{10}{2} \bigg(2(4) + (10 - 1)3 \bigg)S
10
=
2
10
(2(4)+(10−1)3)
\begin{gathered}\rm \: S_{10} = 5 \bigg(8 + 27 \bigg) \\ \end{gathered}
S
10
=5(8+27)
\begin{gathered}\rm \: S_{10} = 5 \times 35 \\ \end{gathered}
S
10
=5×35
\begin{gathered}\rm\implies \:S_{10} = 175 \\ \end{gathered}
⟹S
10
=175