Question

The 6thand 17th term of an ap are 19th and 41 respectively find 30th term

Answer 1

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

Let assume that,

↝ First term of an AP series is a

and

↝ Common difference of an AP series is d.

Given that,

$\rm :\longmapsto\:a_6 = 19 \: \: \: \: and \: \: \: \: a_{17} = 41$

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,

$\rm :\longmapsto\:a_6 = 19$

$\rm :\longmapsto\:a + (6 - 1)d = 19$

$\rm :\longmapsto\:a + 5d = 19$

$\bf\implies \:a = 19 - 5d - - - - (1)$

Also,

$\rm :\longmapsto\:a_{17}= 41$

$\rm :\longmapsto\:a + (17 - 1)d = 41$

$\rm :\longmapsto\:a + 16d = 41$

On substituting the value from equation (1), we get

$\rm :\longmapsto\:19 - 5d + 16d = 41$

$\rm :\longmapsto\:11d = 41 - 19$

$\rm :\longmapsto\:11d = 22$

$\bf\implies \:d = 2$

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

$\rm :\longmapsto\:a = 19 - 5 \times 2$

$\rm :\longmapsto\:a = 19 - 10$

$\bf\implies \:a \: = \: 9$

So,

$\purple{ \bf{ \: \rm :\longmapsto\:a_{30} = a + (30 - 1)d \: }}$

$\purple{ \bf{ \: \rm :\longmapsto\:a_{30} = a + 29d \: }}$

$\purple{ \bf{ \: \rm :\longmapsto\:a_{30} = 9 + 29(2) \: }}$

$\purple{ \bf{ \: \rm :\longmapsto\:a_{30} = 9 + 58 \: }}$

$\purple{ \bf{ \: \rm :\longmapsto\:a_{30} = 67 \: }}$

Additional Information :-

↝ Sum of n  terms of an arithmetic sequence is,

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

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.