The sum of three terms of an AP is 12 The product of first and third term is 8 greater than second term find the termswho will answer correctly I will mark u as brainlist plz don’t spam About the author Arianna
Given that, the sum of three terms of an AP is 12. & The product of first and third term is 8 greater than second term. So, Let’s consider a – d , a , a + d be three terms of A.P. ⠀⠀━━━━━━━━━━━━━━━━━━⠀⠀⠀⠀ [tex]{\underline{\pmb{\frak{\bigstar\:According\:to\:the\:Question\::}}}}\\\\[/tex] Sum of first three terms of AP is 12. ⠀⠀ [tex]\dashrightarrow\sf (a – d) + (a) + (a + d) = 12\\\\\\ \dashrightarrow\sf 3a\: \cancel{- \: d} \: \cancel{+\: d} = 12\\\\\\ \dashrightarrow\sf 3a = 12\\\\\\ \dashrightarrow\sf a = \cancel{\dfrac{12}{3}}\\\\\\ \dashrightarrow{\underline{\boxed{\pmb{\frak{a = 4}}}}}\:\bigstar\\\\[/tex] [tex]\therefore\:{\underline{\sf{Second\:term\:(a)\:of\:AP\:is\:{\pmb{4}}.}}}[/tex] ⠀⠀━━━━━━━━━━━━━━━━━━⠀⠀⠀⠀ ▪︎Now, Let’s solve second Condition — ⠀⠀ The product of first and third term is 8 greater than second term. ⠀⠀ [tex]\qquad\dashrightarrow\sf (a – d)(a + d) = a + 8\\\\\\ \qquad\dashrightarrow\sf (4 – d)(4 + d) = 4 + 8\\\\\\ \qquad\dashrightarrow\sf 4^2 – d^2 = 12\\\\\\ \qquad\dashrightarrow\sf 16 – d^2 = 12\\\\\\ \qquad\dashrightarrow\sf d^2 = 16 – 12\\\\\\ \qquad\dashrightarrow\sf d^2 = 4\\\\\\ \qquad\dashrightarrow{\underline{\boxed{\pmb{\frak{d = 2}}}}}\:\bigstar\\\\[/tex] [tex]\therefore\:{\underline{\sf{Common\:diffrence\:(d)\:between\:the\:terms\:of\:AP\:is\:{\pmb{2}}.}}}[/tex] ⠀⠀━━━━━━━━━━━━━━━━━━⠀⠀⠀⠀ ❍ Therefore, The required terms of A.P. are : ⠀⠀ [tex]\qquad\quad\twoheadrightarrow\sf (a – d) \:,\: a \:,\: (a + d)\\\\\\ \qquad\quad\twoheadrightarrow\sf (4 – 2) \:,\: 4 \:,\: (4 + 2)\\\\\\ \qquad\quad\twoheadrightarrow{\pmb{\frak{\purple{2 \:,\: 4 \:,\: 6}}}}[/tex] Reply
Given :- The sum of three terms of an AP is 12 The product of first and third term is 8 greater than second term To Find :- Terms Solution :- Let the terms of the AP be a – d, a, a + d a – d + a + a + d = 12 (a + a + a) + (d – d) = 12 3a = 12 12/3 = a 4 = a Now (a – d) × (a + d) = a + 8 (4 – d) × (4 + d) = 8 + 4 (4 × 4) – (d × d) = 8 + 4 16 – d² = 8 + 4 -d² = 8 – 12 -d² = -4 d² = 4 d = √4 d = ±2 Finding the term a – d = 4 – 2 = 2 a = 4 a + d = 4 + 2 = 6 [tex]\\[/tex] Reply
Given that, the sum of three terms of an AP is 12. & The product of first and third term is 8 greater than second term.
So, Let’s consider a – d , a , a + d be three terms of A.P.
⠀⠀━━━━━━━━━━━━━━━━━━⠀⠀⠀⠀
[tex]{\underline{\pmb{\frak{\bigstar\:According\:to\:the\:Question\::}}}}\\\\[/tex]
⠀⠀
[tex]\dashrightarrow\sf (a – d) + (a) + (a + d) = 12\\\\\\ \dashrightarrow\sf 3a\: \cancel{- \: d} \: \cancel{+\: d} = 12\\\\\\ \dashrightarrow\sf 3a = 12\\\\\\ \dashrightarrow\sf a = \cancel{\dfrac{12}{3}}\\\\\\ \dashrightarrow{\underline{\boxed{\pmb{\frak{a = 4}}}}}\:\bigstar\\\\[/tex]
[tex]\therefore\:{\underline{\sf{Second\:term\:(a)\:of\:AP\:is\:{\pmb{4}}.}}}[/tex]
⠀⠀━━━━━━━━━━━━━━━━━━⠀⠀⠀⠀
▪︎Now, Let’s solve second Condition —
⠀⠀
⠀⠀
[tex]\qquad\dashrightarrow\sf (a – d)(a + d) = a + 8\\\\\\ \qquad\dashrightarrow\sf (4 – d)(4 + d) = 4 + 8\\\\\\ \qquad\dashrightarrow\sf 4^2 – d^2 = 12\\\\\\ \qquad\dashrightarrow\sf 16 – d^2 = 12\\\\\\ \qquad\dashrightarrow\sf d^2 = 16 – 12\\\\\\ \qquad\dashrightarrow\sf d^2 = 4\\\\\\ \qquad\dashrightarrow{\underline{\boxed{\pmb{\frak{d = 2}}}}}\:\bigstar\\\\[/tex]
[tex]\therefore\:{\underline{\sf{Common\:diffrence\:(d)\:between\:the\:terms\:of\:AP\:is\:{\pmb{2}}.}}}[/tex]
⠀⠀━━━━━━━━━━━━━━━━━━⠀⠀⠀⠀
❍ Therefore, The required terms of A.P. are :
⠀⠀
[tex]\qquad\quad\twoheadrightarrow\sf (a – d) \:,\: a \:,\: (a + d)\\\\\\ \qquad\quad\twoheadrightarrow\sf (4 – 2) \:,\: 4 \:,\: (4 + 2)\\\\\\ \qquad\quad\twoheadrightarrow{\pmb{\frak{\purple{2 \:,\: 4 \:,\: 6}}}}[/tex]
Given :-
The sum of three terms of an AP is 12 The product of first and third term is 8 greater than second term
To Find :-
Terms
Solution :-
Let the terms of the AP be a – d, a, a + d
a – d + a + a + d = 12
(a + a + a) + (d – d) = 12
3a = 12
12/3 = a
4 = a
Now
(a – d) × (a + d) = a + 8
(4 – d) × (4 + d) = 8 + 4
(4 × 4) – (d × d) = 8 + 4
16 – d² = 8 + 4
-d² = 8 – 12
-d² = -4
d² = 4
d = √4
d = ±2
Finding the term
a – d = 4 – 2 = 2
a = 4
a + d = 4 + 2 = 6
[tex]\\[/tex]