Question

The
Sum of three teams of An AP
is 12 The product of first and third
telms is 8 greater than the second
find the
the term .
20m​

Answer 1

Answer:

An acid is a substance that donates protons (in the Brønsted-Lowry definition) or accepts a pair of valence electrons to form a bond (in the Lewis definition).

Step-by-step explanation:

An acid is a substance that donates protons (in the Brønsted-Lowry definition) or accepts a pair of valence electrons to form a bond (in the Lewis definition).

Answer 2

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.

⠀⠀⠀

$\begin{gathered}{\underline{\pmb{\frak{\bigstar\:According\:to\:the\:Question\::}}}}\\\\\end{gathered}$

Sum of first three terms of AP is 12.

⠀⠀

$\begin{gathered}\dashrightarrow\sf (a - d) + (a) + (a + d) = 12\\\\\\ \dashrightarrow\sf 3a\: \cancel{- \: d} \: \cancel{+\: d} = 12\\\\\\ \dashrightarrow\sf 3a = 12\\\\\\ \dashrightarrow\sf a = \cancel{\dfrac{12}{3}}\\\\\\ \dashrightarrow{\underline{\boxed{\pmb{\frak{a = 4}}}}}\:\bigstar\\\\\end{gathered}$

$\therefore\:{\underline{\sf{Second\:term\:(a)\:of\:AP\:is\:{\pmb{4}}.}}}$

⠀⠀⠀⠀

▪︎Now, Let's solve second Condition —

⠀⠀

• The product of first and third term is 8 greater than second term.

⠀⠀

$\begin{gathered}\qquad\dashrightarrow\sf (a - d)(a + d) = a + 8\\\\\\ \qquad\dashrightarrow\sf (4 - d)(4 + d) = 4 + 8\\\\\\ \qquad\dashrightarrow\sf 4^2 - d^2 = 12\\\\\\ \qquad\dashrightarrow\sf 16 - d^2 = 12\\\\\\ \qquad\dashrightarrow\sf d^2 = 16 - 12\\\\\\ \qquad\dashrightarrow\sf d^2 = 4\\\\\\ \qquad\dashrightarrow{\underline{\boxed{\pmb{\frak{d = 2}}}}}\:\bigstar\\\\\end{gathered}$

$\therefore\:{\underline{\sf{Common\:diffrence\:(d)\:between\:the\:terms\:of\:AP\:is\:{\pmb{2}}.}}}$

⠀⠀⠀⠀

❍ Therefore, The required terms of A.P. are :

⠀⠀

$\begin{gathered}\qquad\quad\twoheadrightarrow\sf (a - d) \:,\: a \:,\: (a + d)\\\\\\ \qquad\quad\twoheadrightarrow\sf (4 - 2) \:,\: 4 \:,\: (4 + 2)\\\\\\ \qquad\quad\twoheadrightarrow{\pmb{\frak{\purple{2 \:,\: 4 \:,\: 6}}}}\end{gathered}$