Question

Fifth term of an arithmetic squence is 21 and it's ninth term is 37. (a) what is it's common difference? (b) What is it's first term?​

Answer 1

Given: Fifth term of an arithmetic progression is 21. & the 9th term of the progression is 37.

Need to find: Common difference (d) & first term (a)?

⠀⠀━━━━━━━━━━━━━━━━━━━━━━━━━━━━━⠀⠀⠀

▪︎ Fifth term of the AP —

$:\implies\sf a + 4d = 21\qquad\qquad\qquad\qquad\sf\Bigg\lgroup eq^{n}\;(1)\Bigg\rgroup$

▪︎ Ninth term of the AP —

$:\implies\sf a + 8d = 37\qquad\qquad\qquad\qquad\sf\Bigg\lgroup eq^{n}\;(2)\Bigg\rgroup$

⠀⠀━━━━━━━━━━━━━━━━━━━━━━━━━━━━━⠀⠀⠀

$\;\:\;\underline{\bf{\dag} \:\:\sf{From\;eq^{n}\:(1)\;\&\;eq^{n}\;(2)\;:}}$

$:\implies\sf\quad a + 4d = 21\\\\\\:\implies\sf\quad a + 8d = 37\\\\\\:\implies\sf\quad - 4d = - 16 \\\\\\:\implies\sf\quad d = \cancel\dfrac{-16}{-4}\\\\\\:\implies\quad \underline{\boxed{\pmb{\frak{\pink{d = 4}}}}}\;\bigstar$

$\therefore{\underline{\textsf{Hence, Common difference, (d) of AP is \textbf{4}.}}}$

⠀⠀━━━━━━━━━━━━━━━━━━━━━━━━━━━━━⠀⠀⠀

$\;\:\;\underline{\bf{\dag} \:\;\sf{Now, Using\;eq^{n}\:(2)\; :}}$

$:\implies\sf\quad a + 4d = 21\\\\\\:\implies\sf\quad a + 4(4) = 21\\\\\\:\implies\sf\quad a + 16 = 21\\\\\\:\implies\sf\quad a = 21 - 16\\\\\\:\implies\sf\quad\underline{\boxed{\pmb{\frak{\pink{a = 5}}}}}\;\bigstar$

$\therefore{\underline{\textsf{Hence, first term, (a) of AP is \textbf{5}.}}}$

Answer 2

Answer:

• (a) Common difference of AP is 4.
• (b) First term of AP is 5.

Explanation:

Given information,

Fifth term of an arithmetic sequence is 21 and it's ninth term is 37.

• (a) what is it's common difference?
• (b) What is it's first term?

Here,

• $\sf a_{5}$ = 21
• $\sf a_{9}$ = 37
• a = ?
• d = ?

⚘ Using formula of nth term ::

$\bf{\dag}\:{\boxed{\tt{a_{n} = a + (n - 1)d}}}$

⠀⠀⠀━━━━━━━━━━━━━━━━━━━━━━

(a)

➨ $\tt a_{5} = a + (5 - 1)d$

➨ $\tt 21 = a + 4d$

➨ $\tt a = 21 - 4d\qquad- (1)$

Also,

➨ $\tt a_{9} = a + (9 - 1)d$

➨ $\tt 37 = a + 8d$

➨ $\tt a = 37 - 8d\qquad- (2)$

From (1) & (2) we get,

➨ $\tt 21 - 4d = 37 - 8d$

➨ $\tt 21 - 37 = - 8d + 4d$

➨ $\tt \cancel{-} 16 = \cancel{-} 4d$

➨ $\tt 4d = 16$

➨ $\tt d = {\cancel{\dfrac{16}{4}}}$

➨ d = 4

• Hence, common difference (d) of AP is 4.

(b)

Put d = 4 in (1) we get,

➨ $\tt a = 21 - (4\:\times\:4)$

➨ $\tt a = 21 - 16$

➨ a = 5

• Hence, first term (a) of AP is 5.

Verification:

➨ $\tt a_{5} = a + (5 - 1)d$

➨ $\tt 21 = a + 4d$

By putting value of a and d in above equation we get,

➨ $\tt 21 = 5 + (4\:\times\:4)$

➨ $\tt 21 = 5 + 16$

➨ $\tt 21 = 21$

➨ LHS = RHS

Also,

➨ $\tt a_{9} = a + (9 - 1)d$

➨ $\tt 37 = a + 8d$

By putting value of a and d in above equation we get,

➨ $\tt 37 = 5 + (8\:\times\:4)$

➨ $\tt 37 = 5 + 32$

➨ $\tt 37 = 37$

➨ LHS = RHS

Hence, Verified ✔

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬