Question

four numbers are in AP the sum of first and fourth is 14 and product of second and third is 45 find the number ​

Answer 1

⭐Given:⭐

$\tt Sum \: of \: 1_{st}\: and \: 4_{th}\: number \: in\: AP \: = 14$

$\tt Product\: of \: 2_{nd}\: and \: 3_{rd} \: number \: in \: AP \: = 45$

⭐Answer:⭐

Explanation

Let the four numbers be (a-3d) , (a-d) , (a+d) and (a+3d) respectively.

ATQ,

$(a - 3d) + (a + 3d) = 14$

$2a = 14$

$a = 7$

ATQ,

$(a - d) \times (a + d) = 45$

We know that,

$(a + b) \times (a - b) = ( {a}^{2} - {b}^{2} )$

${a}^{2} - {d}^{2} = 45$

Substituting the value of a in the above equation, we get:

$49 - {d}^{2} = 45$

${d}^{2} = 4$

$d = 2$

$\bf \: \therefore \: first \: term \: = 7$

$\bf \: \therefore \: common \: difference = 2$

Numbers are:

1) (a-3d) = 7-3*2 = 1

2) (a-d) = 7-2 = 5

3) (a+d) = 7+2 = 9

4) (a+3d) = 7+3*2 = 13

⭐More About AP⭐

nth term of an AP formulas

$\sf 1) \: n_{th} \: term \: of \: any \: AP \: = a + (n - 1)d$

$\sf 2) \: (i) \: n_{th} \: term \: from \: the \: end \: of \: an \: AP \: = a + (m - n)d$

$\sf 2) \: (ii) \: n_{th} \: term \: from \: the \: end \: of \: an \: AP = l - (n - 1)d$

$\sf 3) \: Difference \: of \: two \: terms = (m - n)d$

where m and n is the position of the term in AP

$\tt 4) \: Middle \: term \: of \: a \: finite \: AP$

$\sf If\:n\:is\:odd=(\frac{n + 1}{2})th\:term$

$\sf If \: n \: is \: even =\frac{n}{2} \: th\: term \: and \: (\frac{\:n\:}{\:2\:} + 1\:)th\:term$

Sum Formulas

$\sf Sum \: of \: first \: n \: terms \: of \: an \:AP = \frac{n}{2} [ \: 2a + (n - 1)d \: ]$

$\sf Sum \: of \: first \: n \: natural \: numbers = \frac{n(n + 1)}{2}$

$\sf Sum \: of \: AP \: having \: last \: term = \frac{n}{2} [ \: a + l \: ]$

⭐Numbers in an AP⭐

Three Numbers in an AP:

(a-d) , a and (a+d)

Four Numbers in an AP:

(a-3d) , (a-d) , (a+d) and (a+3d)

Five Numbers in an AP:

(a-2d) , (a-d) , a , (a+d) and (a+2d)