Question

2.Consider an arithmetic sequence whose 6th term is 40 and 9th term is 58. (a) Find its common difference (b)Find out its first term (c) Write down the algebraic expression (d) Check whether 150 is a term or not

Answer 1

$\large\underline{\sf{Solution-}}$

Wᴇ ᴋɴᴏᴡ ᴛʜᴀᴛ,

↝ nᵗʰ term of an arithmetic sequence is,

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

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,

According to statement,

$\rm :\longmapsto\:a_6 = 40$

$\rm :\longmapsto\:a + (6 - 1)d = 40$

$\rm :\longmapsto\:a + 5d = 40$

$\bf\implies \:a = 40 - 5d - - - (1)$

Also, given that

$\rm :\longmapsto\:a_9 = 58$

$\rm :\longmapsto\:a + (9 - 1)d = 58$

$\rm :\longmapsto\:a + 8d = 58$

$\rm :\longmapsto\:40 - 5d+ 8d = 58 \: \: \: \: \: \: \{using \: (1) \}$

$\rm :\longmapsto\:40 + 3d = 58$

$\rm :\longmapsto\:3d = 58 - 40$

$\rm :\longmapsto\:3d = 18$

$\bf\implies \:d = 6$

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

$\rm :\longmapsto\:a + 5 \times 6 = 40$

$\rm :\longmapsto\:a + 30 = 40$

$\rm :\longmapsto\:a = 40 - 30$

$\bf\implies \:a = 10$

Hence,

↝ First term of an AP, a = 10

and

↝ Common difference of an AP, d = 6

Now,

Wᴇ ᴋɴᴏᴡ ᴛʜᴀᴛ,

↝ nᵗʰ term of an arithmetic sequence is,

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

So, on substituting the values, we get

$\rm :\longmapsto\:a_n = a + (n - 1)d$

$\rm :\longmapsto\:a_n = 10 + (n - 1)6$

$\rm :\longmapsto\:a_n = 10 + 6n - 6$

$\bf:\longmapsto\:a_n = 6n + 4$

↝ To check whether 150 is the term of given series or not.

↝ Let assume that nth term of the series is 150. Now we find the value of n. If value of 'n' is a natural number, then 150 is the term of series otherwise its not a term of series.

So, we have now

$\red{\rm :\longmapsto\:a = 10}$

$\red{\rm :\longmapsto\:d = 6}$

$\red{\rm :\longmapsto\:a_n = 150}$

Now, using

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

On substituting the values, we get

$\rm :\longmapsto\:150 = 10 + (n - 1)6$

$\rm :\longmapsto\:150 - 10 = 6n - 6$

$\rm :\longmapsto\:140 = 6n - 6$

$\rm :\longmapsto\:140 + 6 = 6n$

$\rm :\longmapsto\:6n = 146$

$\bf\implies \:n = \dfrac{146}{6} = \dfrac{73}{3} \: \: \cancel \in \: natural \: number$

↝  Hence, 150 is not the term of given AP.