find the measure of X in each of the following
Answers
Step-by-step explanation:
\large\underline{\sf{Solution-}}
Solution−
Given that,
An AP consists of 52 terms of which 3rd term is 13 and the last term is 106.
Let assume that, first term of an AP is a and common difference of an AP is d.
So, we have
\begin{gathered}\rm \: n \: = \: 52 \\ \end{gathered}
n=52
\begin{gathered}\rm \: a_3 \: = 13 \\ \end{gathered}
a
3
=13
\begin{gathered}\rm \: a_{52} \: = 106 \\ \end{gathered}
a
52
=106
Wᴇ ᴋɴᴏᴡ ᴛʜᴀᴛ,
↝ nᵗʰ term of an arithmetic sequence is,
\begin{gathered}\begin{gathered}\:\:{\underline{{\boxed{\bf{{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 sequence.
n is the no. of terms.
d is the common difference.
Tʜᴜs,
\begin{gathered}\rm \: a_3 \: = 13 \\ \end{gathered}
a
3
=13
\begin{gathered}\rm \: a + (3 - 1)d = 13 \\ \end{gathered}
a+(3−1)d=13
\begin{gathered}\rm\implies \:a + 2d = 13 - - - (1) \\ \end{gathered}
⟹a+2d=13−−−(1)
Also,
\begin{gathered}\rm \: a_{52} \: = 106 \\ \end{gathered}
a
52
=106
\begin{gathered}\rm \: a + (52 - 1)d = 106 \\ \end{gathered}
a+(52−1)d=106
\begin{gathered}\rm\implies \:\rm \: a + 51d = 106 - - - (2) \\ \end{gathered}
⟹a+51d=106−−−(2)
On Subtracting equation (1) from equation (2), we get
\begin{gathered}\rm \: 49d = 93 \\ \end{gathered}
49d=93
\begin{gathered}\rm\implies \:d \: = \: \dfrac{93}{49} \\ \end{gathered}
⟹d=
49
93
On substituting the value of d in equation (1), we get
\begin{gathered}\rm \: a \: + \: 2 \times \dfrac{93}{49} = 13 \\ \end{gathered}
a+2×
49
93
=13
\begin{gathered}\rm \: a \: + \: \dfrac{186}{49} = 13 \\ \end{gathered}
a+
49
186
=13
\begin{gathered}\rm \: a \: = \: 13 - \dfrac{186}{49} \\ \end{gathered}
a=13−
49
186
\begin{gathered}\rm \: a \: = \: \dfrac{637 - 186}{49} \\ \end{gathered}
a=
49
637−186
\begin{gathered}\rm\implies \:\rm \: a \: = \: \dfrac{451}{49} \\ \end{gathered}
⟹a=
49
451
Now, Consider
\begin{gathered}\rm \: a_{32} \\ \end{gathered}
a
32
\begin{gathered}\rm \: = \: a + (32 - 1)d \\ \end{gathered}
= a+(32−1)d
\begin{gathered}\rm \: = \: a + 31d \\ \end{gathered}
= a+31d
On substituting the values of a and d, we get
\rm \: = \: \dfrac{451}{49} + 31 \times \dfrac{93}{49}=
49
451
+31×
49
93
\rm \: = \: \dfrac{451}{49} + \dfrac{2883}{49}=
49
451
+
49
2883
\begin{gathered}\rm \: = \: \dfrac{451 + 2883}{49} \\ \end{gathered}
=
49
451+2883
\begin{gathered}\rm \: = \: \dfrac{3334}{49} \\ \end{gathered}
=
49
3334
\begin{gathered}\rm\implies \:\boxed{\sf{ \:\rm \:a_{32} = \: \dfrac{3334}{49} \: \: }} \\ \end{gathered}
⟹
a
32
=
49
3334
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Additional Information :-
↝ Sum of n terms of an arithmetic sequence is,
\begin{gathered}\begin{gathered}\:\:{\underline{{\boxed{\bf{{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 sequence.
n is the no. of terms.
d is the common difference.