Math, asked by llAksharall7, 23 days ago

An AP consists of 52 terms of which 3rd term is 13 and the last term is 106.Find the 32th term of the AP.

Need answer with full explanation!!​

Answers

Answered by mathdude500
26

\large\underline{\sf{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

\rm \: n \:  =  \: 52 \\

\rm \: a_3 \:  = 13 \\

\rm \: a_{52} \:  = 106 \\

Wᴇ ᴋɴᴏᴡ ᴛʜᴀᴛ,

↝ nᵗʰ term of an arithmetic sequence is,

\begin{gathered}\:\:{\underline{{\boxed{\bf{{a_n\:=\:a\:+\:(n\:-\:1)\:d}}}}}} \\ \end{gathered}

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,

\rm \: a_3 \:  = 13 \\

\rm \: a + (3 - 1)d = 13 \\

\rm\implies \:a + 2d = 13 -  -  - (1) \\

Also,

\rm \: a_{52} \:  = 106 \\

\rm \: a + (52 - 1)d = 106 \\

\rm\implies \:\rm \: a + 51d = 106 -  -  - (2) \\

On Subtracting equation (1) from equation (2), we get

\rm \: 49d = 93 \\

\rm\implies \:d \:  =  \: \dfrac{93}{49}  \\

On substituting the value of d in equation (1), we get

\rm \: a \:  +  \: 2 \times \dfrac{93}{49}  = 13 \\

\rm \: a \:  +  \:  \dfrac{186}{49}  = 13 \\

\rm \: a \:   =   \: 13 -  \dfrac{186}{49} \\

\rm \: a \:   =   \: \dfrac{637 - 186}{49} \\

\rm\implies \:\rm \: a \:   =   \: \dfrac{451}{49} \\

Now, Consider

\rm \: a_{32} \\

\rm \: =  \: a + (32 - 1)d \\

\rm \: =  \: a + 31d \\

On substituting the values of a and d, we get

\rm \: =  \: \dfrac{451}{49} + 31 \times \dfrac{93}{49}

\rm \: =  \: \dfrac{451}{49} +  \dfrac{2883}{49}

\rm \: =  \: \dfrac{451 + 2883}{49}  \\

\rm \: =  \: \dfrac{3334}{49}  \\

\rm\implies \:\boxed{\sf{  \:\rm \:a_{32} =  \: \dfrac{3334}{49}  \:  \: }} \\

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Additional Information :-

↝ Sum of n  terms of an arithmetic sequence is,

\begin{gathered}\:\:{\underline{{\boxed{\bf{{S_n\:=\dfrac{n}{2} \bigg(2 \:a\:+\:(n\:-\:1)\:d \bigg)}}}}}} \\ \end{gathered}

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.

Answered by s8a1583aritra1756
11

Answer:

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

Wᴇ ᴋɴᴏᴡ ᴛʜᴀᴛ,

↝ nᵗʰ term of an arithmetic sequence is,

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,

Also,

On Subtracting equation (1) from equation (2), we get

On substituting the value of d in equation (1), we get

Now, Consider

On substituting the values of a and d, we get

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Additional Information :-

↝ Sum of n  terms of an arithmetic sequence is,

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.

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