Question

In an arithmetic sequence, 8th term is 17 and 20th term is 41. Find a) the common difference b) the first term c) 35 th term of the sequence *

Answer 1

$\bf\underline \blue {To \: \mathscr{f}ind:-}$

• a) the common difference
• b) the first term
• c) 35 th term of the sequence

$\huge\bf\underline \red{ \mathbb{S}olution...}$

$\bf\underline{\purple{Given:-}}$

• 8th term = 17
• 20th term = 41

• an = a + (n - 1)d

• 8th term can be written as :-

⇝a8 = a + (8-1)d

⇝a8 = a + 7d

⇝ a + 7d = 17 .....1)

• 20th term can be written as:-

⇝a20 = a +(20 - 1) d

⇝a20 = a + 19d

⇝ a + 19d = 41 .....2)

• Solving eq ...1) and ...2)

⠀⠀⠀⠀a + 19d = 41

⠀⠀⠀⠀a + 7d = 17

⠀⠀⠀⠀--⠀⠀--⠀ ⠀--

⠀⠀⠀⠀⠀⠀⠀12d = 24

⠀⠀⠀⠀⠀⠀⠀⠀d = 2

Putting value of d in ...1)

⇛a + 7d = 17

⇛ a + 7 × 2 = 17

⇛ a + 14 = 17

⇛ a = 17 - 14

⇛ a = 3

Now,

• 35th term of sequence :-

⇛ a35 = a + 34d

⇛ a35 = 3 + 34 × 2

⇛ a35 = 3 + 68

⇛ a35 = 71

So,

• a) the common difference (d) = 2
• b) the first term (a) = 3
• c) 35 th term of the sequence = 71

━━━━━━━━━━━━━━━━━━━━━━━━━

Answer 2

$\green{\bf{\underline{\underline{Answer:-}}}}$

$\blue{\bold{\underline{\underline{Given:}}}}$.

• 8th Term = 17
• 20th Term = 41

$\orange{\mathcal{\underline{\underline{To \: find:}}}}$

• A) Common Difference

• B) The 1st Term

• C) 35th Term

$\small \underline{ \blue{ \boxed{ \bf \red{a _{n} = a + (n - 1)d}}}}$

8th Term is

$a _{8} = a +( 8 - 1)d$

$a + 7d = 17 \: \: \: \: ....(i)$

20th Term is

$a _{20} = a + (20 - 1)d$

$a + 19d = 41 \: \: \: \: ....(ii)$

{Subtract(i)from(ii)}

$12d = 24$

$d = 2$

Putting The Value Of d in (i) :

$a + (7 \times 2) = 17$

$a = 3$

Thus,35th Term is

$= a + 34d$

$= 3 +( 34 \times 2)$

$a _{35} = 71$

A) d = 2

B) a = 3

C) a35 = 71

$\bf\blue{Hope\ it\ helps.}$

$\bf\pink{Plz\ Mark\ As\ Brainliest.}$