Question

8th and
18th term of an arithmetic sequence
is

67 and 147.

What
is the first term?​

Answer 1

Answer:

$\bigstar{\bold{First\:term(a_1)=11}}$

Step-by-step explanation:

$\Large{\underline{\underline{\bf{Given:}}}}$

• 8th term (a₈) = 67
• 18th term (a₁₈) = 147

$\Large{\underline{\underline{\bf{To\:Find:}}}}$

• First term of the A.P (a₁)

$\Large{\underline{\underline{\bf{Solution:}}}}$

→ First we need to find the common difference(d) of the A.P

  $d=\dfrac{a_m-a_n}{m-n}$

where $a_m$ = 147, $a_n$ = 67, m = 18, n = 8

→ Substituting the given data we get,

$d=\dfrac{147-67}{18-8}$

$d=\dfrac{80}{10}$

$d=8$

→ The eighth term of an A.P can be written as

a₈ = a₁ + 7d

→ Substitute the value of a₈ and d

67 = a₁ + 7 × 8

a₁ = 67 - 56

$\boxed{\bold{First\:term(a_1)=11}}$

$\Large{\underline{\underline{\bf{Notes:}}}}$

→ The common difference of an A.P is the difference between its two consecutive terms.

$d=a_2-a_1$

$d=\dfrac{a_m-a_n}{m-n}$

Answer 2

$\bf\huge\star\underline{\underline{ANSWER}}\ \star$

$\rm Given \ ,\\\\8^{th} \ term = 67\\\\18^{th} \ term = 147$

____________________________________

     $\bf\boxed{\boxed{\bf Commom \ difference = \dfrac{Term \ difference }{Position \ difference}}}$

$\rm\longrightarrow Common \ difference \ , d =\dfrac{147-67}{18-8}$

                                       $= \rm \dfrac{80}{10}$

                                       $= \bf 8$

_____________________________________

$\rm\longrightarrow 1^{st} \ term = 8^{th} \ term -7d\\\\\longrightarrow 1^{st} \ term = 67-7\times8\\\\\longrightarrow 1^{st} \ term = 67-56\\\\\longrightarrow\bf\underline{ 1^{st} \ term = 11}$______________________________________

HOPE IT HELPS U ..........  :)