Question

Consider an arithmetic sequence whose 6 th term is 40 and 9 th term is 58 (¡) Find 25 th term of the sequenc

Answer 1

$\large{\underline{\underline{\frak{\;\;AnSwer\;:}}}}$

• We're given with two terms of an AP (Arithmetic Progression). The 6th term of the AP is 40 and the 9th term of the AP is 58. Here, we're asked to calculate the 25th term of the AP. For that, we've to find out the Common difference (d), and first term (a) of the AP. Therefore:

$:\implies\sf a + 5d = 40\qquad\qquad\qquad\sf\Bigg\lgroup\ eq^{n}\:(i)\Bigg\rgroup$

Similarly,

$:\implies\sf a + 8d = 58\qquad\qquad\qquad\sf\Bigg\lgroup\ eq^{n}\:(ii)\Bigg\rgroup$

$\;\:\;\underline{\bf{\dag} \:\sf{From\;eq^{n}\;(i)\;\&\;eq^{n}\;{(ii)}\: :}}$⠀

$\dashrightarrow\sf\quad a + 5d = 40$

$\dashrightarrow\sf\quad a + 8d = 58$

$\dashrightarrow\sf\quad 3d = 18$

$\dashrightarrow\sf\quad d = \cancel\dfrac{18}{3}$

$\dashrightarrow\quad\underline{\boxed{\pmb{\sf{d = 6}}}}$

$\;\;\;\underline{\bf{\dag} \:\sf{Finding\;First\;term\;(a) \;of\;AP\: :}}$⠀⠀⠀⠀

$\dashrightarrow\sf\quad a + 5d = 40$

$\dashrightarrow\sf\quad a + 5(6) = 40$

$\dashrightarrow\sf\quad a + 30 = 40$

$\dashrightarrow\sf\quad a = 40 - 30$

$\dashrightarrow\quad\underline{\boxed{\pmb{\sf{a = 10}}}}$

⠀$\underline{\bf{\dag} \:\sf{Finding\;25^{th}\;term \;of\;AP\: :}}$⠀⠀⠀⠀

$\dashrightarrow\sf\quad a_n = a + (n - 1)d$

$\dashrightarrow\sf\quad a_{25} = 10 + (25 - 1)6$

$\dashrightarrow\sf\quad a_{25} = 10 + 24 \times 6$

$\dashrightarrow\sf\quad a_{25} = 10 + 144$

$\dashrightarrow\quad\underline{\boxed{\pmb{\sf{a_{25} = 154}}}}\:\bigstar$

$\therefore{\underline{\textsf{Hence, the 25th term of the AP is \textbf{154}.}}}$

⠀⠀⠀⠀⠀⠀

Answer 2

Answer:

$\large{\underline{\underline{\frak{\;\;Question\;:}}}}$

• Consider An arithmetic sequence whose 6th term of an AP is 40 and 9th term of an AP is 58.Find 25th term of an AP.

$\large{\underline{\underline{\frak{\;\;Answer\;:}}}}$

• 25th term of an AP is 154.

$\large{\underline{\underline{\frak{\;\;Given\;:}}}}$

• An arithmetic sequence whose 6th term of an AP is 40 and 9th term of an AP is 58.

$\large{\underline{\underline{\frak{\;\;To find\;:}}}}$

• The 25 th term of an AP.

$\large{\underline{\underline{\frak{\;\;Solution\;:}}}}$

Case (i)

• In the question given that An arithmetic sequence whose 6th term of an AP is 40.
• we can write it as,

$a + 5d = 40$

Case (ii)

• In the question given that An arithmetic sequence whose 9th term of an AP is 58.

• We can write it as,

$a + 8d = 58$

• By solving equation 1 and 2 we get that,

• a+5d=40
• a+8d=58
• -3d=-18
• $d = \frac{ - 18}{ - 3} = \frac{18}{3} = 6$

Now,

• Finding first term we get that,

• a+8d=58
• a+8 (6)=58
• a+48=58
• a=58-48
• a=10.

Hence,

• First term here in this question is 10.

Now,

• According to the given question we should find 25th term of an AP

• a25=a+24d
• a25=10+24 (6)
• a25=10+144
• a25=154.

Therefore ,

• The 25th term of an AP is 154.

Hope it helps u mate .

Thank you .