Question

Х in ABC X 90 X X B C​

Answer 1

Answer:

$\large\underline{\sf{Solution-}}Solution−</p><p></p><p>Given that,</p><p></p><p>An AP consists of 52 terms of which 3rd term is 13 and the last term is 106.</p><p></p><p>Let assume that, first term of an AP is a and common difference of an AP is d.</p><p></p><p>So, we have</p><p></p><p>\begin{gathered}\rm \: n \: = \: 52 \\ \end{gathered}n=52</p><p></p><p>\begin{gathered}\rm \: a_3 \: = 13 \\ \end{gathered}a3=13</p><p></p><p>\begin{gathered}\rm \: a_{52} \: = 106 \\ \end{gathered}a52=106</p><p></p><p>Wᴇ ᴋɴᴏᴡ ᴛʜᴀᴛ,</p><p></p><p>↝ nᵗʰ term of an arithmetic sequence is,</p><p></p><p>\begin{gathered}\begin{gathered}\:\:{\underline{{\boxed{\bf{{a_n\:=\:a\:+\:(n\:-\:1)\:d}}}}}} \\ \end{gathered}\end{gathered}an=a+(n−1)d</p><p></p><p>Wʜᴇʀᴇ,</p><p></p><p>aₙ is the nᵗʰ term.</p><p></p><p>a is the first term of the sequence.</p><p></p><p>n is the no. of terms.</p><p></p><p>d is the common difference.</p><p></p><p>Tʜᴜs,</p><p></p><p>\begin{gathered}\rm \: a_3 \: = 13 \\ \end{gathered}a3=13</p><p></p><p>\begin{gathered}\rm \: a + (3 - 1)d = 13 \\ \end{gathered}a+(3−1)d=13</p><p></p><p>\begin{gathered}\rm\implies \:a + 2d = 13 - - - (1) \\ \end{gathered}⟹a+2d=13−−−(1)</p><p></p><p>Also,</p><p></p><p>\begin{gathered}\rm \: a_{52} \: = 106 \\ \end{gathered}a52=106</p><p></p><p>\begin{gathered}\rm \: a + (52 - 1)d = 106 \\ \end{gathered}a+(52−1)d=106</p><p></p><p>\begin{gathered}\rm\implies \:\rm \: a + 51d = 106 - - - (2) \\ \end{gathered}⟹a+51d=106−−−(2)</p><p></p><p>On Subtracting equation (1) from equation (2), we get</p><p></p><p>\begin{gathered}\rm \: 49d = 93 \\ \end{gathered}49d=93</p><p></p><p>\begin{gathered}\rm\implies \:d \: = \: \dfrac{93}{49} \\ \end{gathered}⟹d=4993</p><p></p><p>On substituting the value of d in equation (1), we get</p><p></p><p>\begin{gathered}\rm \: a \: + \: 2 \times \dfrac{93}{49} = 13 \\ \end{gathered}a+2×4993=13</p><p></p><p>\begin{gathered}\rm \: a \: + \: \dfrac{186}{49} = 13 \\ \end{gathered}a+49186=13</p><p></p><p>\begin{gathered}\rm \: a \: = \: 13 - \dfrac{186}{49} \\ \end{gathered}a=13−49186</p><p></p><p>\begin{gathered}\rm \: a \: = \: \dfrac{637 - 186}{49} \\ \end{gathered}a=49637−186</p><p></p><p>\begin{gathered}\rm\implies \:\rm \: a \: = \: \dfrac{451}{49} \\ \end{gathered}⟹a=49451</p><p></p><p>Now, Consider</p><p></p><p>\begin{gathered}\rm \: a_{32} \\ \end{gathered}a32</p><p></p><p>\begin{gathered}\rm \: =  \: a + (32 - 1)d \\ \end{gathered}= a+(32−1)d</p><p></p><p>\begin{gathered}\rm \: =  \: a + 31d \\ \end{gathered}= a+31d</p><p></p><p>On substituting the values of a and d, we get</p><p></p><p>\rm \: =  \: \dfrac{451}{49} + 31 \times \dfrac{93}{49}= 49451+31×4993</p><p></p><p>\rm \: =  \: \dfrac{451}{49} + \dfrac{2883}{49}= 49451+492883</p><p></p><p>\begin{gathered}\rm \: =  \: \dfrac{451 + 2883}{49} \\ \end{gathered}= 49451+2883</p><p></p><p>\begin{gathered}\rm \: =  \: \dfrac{3334}{49} \\ \end{gathered}= 493334</p><p></p><p>\begin{gathered}\rm\implies \:\boxed{\sf{  \:\rm \:a_{32} =  \: \dfrac{3334}{49} \: \: }} \\ \end{gathered}⟹ a32= 493334</p><p></p><p>▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬</p><p></p><p>Additional Information :-</p><p></p><p>↝ Sum of n  terms of an arithmetic sequence is,</p><p></p><p>\begin{gathered}\begin{gathered}\:\:{\underline{{\boxed{\bf{{S_n\:=\dfrac{n}{2} \bigg(2 \:a\:+\:(n\:-\:1)\:d \bigg)}}}}}} \\ \end{gathered}\end{gathered}Sn=2n(2a+(n−1)d)</p><p></p><p>Wʜᴇʀᴇ,</p><p></p><p>Sₙ is the sum of n terms of AP.</p><p></p><p>a is the first term of the sequence.</p><p></p><p>n is the no. of terms.</p><p></p><p>d is the common difference.</p><p></p><p>$