Question

find the 35th term of the A.P 20, 17, 14, 11....​

Answer 1

AnSwer :

• First term (a₁) = 20
• Second term (a₂) = 17
• Third term (a₃) = 14
• Fourth term (a₄) = 11

We need to find the 35ᵗʰ term of an given A. P

• First of all we need to find the common difference of an given A.P and then after finding common difference we can find the 35ᵗʰ term of the A.P.

$\qquad\dag\:\underline{\textbf{Common Difference (d) : }}$

$:\implies \sf Common \: Difference = Second \: term - First \: term$

$:\implies \sf d = a_{2} - a_{1}$

$:\implies \sf d = 17 - 20$

$:\implies \underline{ \boxed{ \frak{ d = - 3}}}$

$\qquad\dag\:\underline{\bf{ {35}^{th} \: term \: of \: the \: A.P: }}$

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

Where,

• aₙ = nᵗʰ term of the A.P
• a = First term of the A.P
• d = Common Difference

$\dashrightarrow\:\:\sf a_{35} = 20 + (35- 1) \times( - 3)$

$\dashrightarrow\:\:\sf a_{35} = 20 + 34 \times( - 3)$

$\dashrightarrow\:\:\sf a_{35} = 20 + ( - 102)$

$\dashrightarrow\:\:\sf a_{35} = 20 - 102$

$\dashrightarrow\:\: \underline{ \boxed{\frak{a_{35} = - 82 }}}$

Answer 2

AnswEr-:

• $\underline{\boxed{\frak { \:a_{35} \:or\:35^{th} \:term\:of\:an\:A.P. \: =\: -82 \:}}}$

$\sf{\dag{\underline {Explanation \:of\:AnswEr \: -:}}}$

• $\sf{\underline {Given-:}}$

• A.P = 20 , 17 , 14 , 11 ....

• $\sf{\underline {Here-:}}$

• $\sf {First\:Term\:of\:A.P \; or \:(a_{1}) = 20}$

• $\sf {Second \:Term\:of\:A.P \; or \:(a_{2}) = 17}$

• $\sf {Third \:Term\:of\:A.P \; or \:(a_{3}) = 14}$

• $\sf {Fourth \:Term\:of\:A.P \; or \:(a_{4}) = 11}$

• $\sf{\underline {To\:Find-:}}$

• The $\sf{35^{th}}$ term of A.P .

$\sf{\dag{\underline {Solution\:of\:Question\: -:}}}$

• $\underbrace{\sf{\underline {Understanding \:the\:Concept \: }}}$

• We , have to Find the $\sf{35^{th}}$ term of A.P .

• For Finding this first we have to Find the common differences or (d) of an Given AP .

• $\sf{\dag{\underline {Common\:\:Differences \: or\: (d) \:-:}}}$

• $\sf{\underline {Common \:Difference \:\:=\: Second \:Term \:\: - First \:Term\:}}$

• $\sf{\underline {Or,}}$

• $\sf{\underline { \:D \:\:=\: a_{2} \:\: - a_{1}\:}}$

• $\sf{\underline {Here-:}}$

• $\sf {Second \:Term\:of\:A.P \; or \:(a_{2}) = 17}$

• $\sf {First\:Term\:of\:A.P \; or \:(a_{1}) = 20}$

• $\sf{\underline {Then,}}$

• $\longrightarrow{\sf { \:D \:\:=\: 17 \:\: - 20\:}}$

• $\longrightarrow{\sf { \:D \:\:=\: -3\:}}$

• $\sf{\underline {Therefore,}}$

• $\boxed{\sf { \:D \:or\:Common \:Difference\:=\: -3\:}}$

$\sf{\bigstar {Now,}}$

$\sf{\dag{\underline {35^{th}\:term \:of\:An\:A.P\: -:}}}$

• $\sf{\underline { \:a_{n}\:=\: a + ( n -1 ) \: d \:}}$

• $\sf{\underline {Here-:}}$

• $\sf {First\:Term\:of\:A.P \; or \:(a) = 20}$

• $\sf {a_{n} = \:n^{th}\:term\:of\;an\:A.P= 35}$

• $\sf { \:D \:or\:Common \:Difference\:=\: -3\:}$

• $\sf{\underline {Now,\:By\:Putting\:known\:values-:}}$

• $\longrightarrow{\sf { \:a_{35} \:\:=\: 20 + ( 35 - 1) \times (-3) \:}}$

• $\longrightarrow{\sf { \:a_{35} \:\:=\: 20 + ( 34) \times (-3) \:}}$

• $\longrightarrow{\sf { \:a_{35} \:\:=\: 20 + (-102) \:}}$

• $\longrightarrow{\sf { \:a_{35} \:\:=\: 20 -102 \:}}$

• $\longrightarrow{\sf { \:a_{35} \:\:=\: -82 \:}}$

$\sf{\dag{\underline {Hence \: -:}}}$

• $\underline{\boxed{\frak { \:a_{35} \:or\:35^{th} \:term\:of\:an\:A.P. \: =\: -82 \:}}}$

_______________________♡______________________