Question

Find the Three numbers in an AP whose sum and product are 15 and 80 respectively??????​

Answer 1

⭐Given:⭐

$\tt Sum \: of \: three \: numbers \: in \: an \: AP \: = \: 15$

$\tt Product \: of \: three \: numbers \: in \: an \: AP \: = \: 80$

⭐Answer:⭐

Explanation:

Let the three numbers in the AP be (a-d) , a , (a+d).

ATQ,

$(a - d) + a + (a + d) = 15$

$a + a + a + d - d = 15$

$3a = 15$

$a = 5$

$\therefore \: first \: term \: = \: 5$

ATQ,

$(a - d) \times a \times (a + d) = 80$

We know that,

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

$( {a}^{2} - {d}^{2} ) \times a = 80$

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

$(25 - {d}^{2} ) \times 5 = 80$

$125 - 5 {d}^{2} = 80$

$- 5 {d}^{2} = 80 - 125$

$- 5 {d}^{2} = - 45$

Dividing both sides by -5 , we get:

${d}^{2} = 9$

Square rooting both sides, we get:

$d = 3$

$\therefore \: common \: difference \: = 3$

First term = 5

Common Difference = 3

Three numbers are:

1) a-d = 5-3 = 2

2) a = 5

3) a+d = 5+3 = 8

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⭐Other AP Formulas:⭐

nth term of an AP formulas

$\sf1) \: n_{th} \: term \: of \: any \: AP \: = a + (n - 1)d$

$\sf2) \: (i) \: n_{th} \: term \: from \: the \: end \: of \: an \: AP \: = a + (m - n)d$

$\sf2) \: (ii) \: n_{th} \: term \: from \: the \: end \: of \: an \: AP = l - (n - 1)d$

$\sf3) \: Difference \: of \: two \: terms = (m - n)d$

$\bf{4) \: Middle \: term \: of \: a \: finite \: AP}$

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

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

Sum Formulas

$\sf1) \: Sum \: of \: first \: n \: terms \: of \: an \:AP = \frac{n}{2} [ \: 2a + (n - 1)d \: ]$

$\sf2) \: Sum \: of \: first \: n \: natural \: numbers = \frac{n(n + 1)}{2}$

$\sf3) \: Sum \: of \: AP \: having \: last \: term \: as \: l = \frac{n}{2} [ \: a + l \: ]$

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Answer 2

GivEn:

• Sum of three terms of an AP = 15

• Product of three terms of an AP = 80

To find:

• Find these three terms of AP.

SoluTion:

☯ Let (a - d), a, (a + d) be the three terms of an AP.

⠀⠀⠀⠀⠀⠀⠀

$\underline{\bigstar\:\boldsymbol{As\:per\:given\:question\::}}$

$\star\;{\underline{\sf{\blue{Sum\; of\; the\; terms = 15}}}}$

$:\implies\sf (a - d) + a + (a + d) = 15\\\\:\implies\sf 3a = 15\\\\:\implies\sf a = \cancel{ \dfrac{15}{3}}\\\\:\implies{\underline{\boxed{\sf{\purple{a = 5}}}}}\;\bigstar$

$\star\;{\underline{\sf{\blue{Product\; of\; the\; terms = 80}}}}$

$:\implies\sf (a - d) a (a + d) = 80\\\\:\implies\sf (a^2 - d^2) a = 80\qquad\bigg\lgroup\bf (a - d)(a + d) = a^2 - d^2 \bigg\rgroup$

$\small\sf\dag\; \underline{Putting\;value\;a\;:}\\\\:\implies\sf (5^2 - d^2)5 = 80\\\\:\implies\sf (25 - d^2) = \cancel{ \dfrac{80}{5}}\\\\:\implies\sf d^2 = 25 - 16\\\\:\implies\sf d^2 = 9\\\\ \small\sf\dag\; \underline{Taking\;sqrt.\;both\;sides\;:}\\\\:\implies\sf \sqrt{d^2} = \sqrt{9}\\\\:\implies{\underline{\boxed{\sf{\purple{d \pm 3}}}}}\;\bigstar$

$\rule{150}2$

Now, we have,

$\sf\;\;\bullet\; a = 5\;;\; d \pm 3$

$\star\;{\underline{\sf{\pink{Therefore,\; required\;three\;terms\;are\;:}}}}$

$\sf\;\;\bullet\; (a - d) = 5 - (+3) \;; \;5 - (-3) = 5 - 3 = \bf{2\;;\;8}\\\\\sf\;\;\bullet\; a = 5\\\\\sf\;\;\bullet\; (a + 5) = 5 + (+3)\;;\;5 + (-3) = \bf{8\;;\;2}$

$\sf\dag\;{\underline{\sf{\purple{Hence,\;the\; required\;terms\;of\;AP\;may\;be\;(2 ,5 ,8)\;or\;(8 ,5, 2).}}}}$

$\rule{150}2$

★ Formula Related to AP (Arithmetic Progression):

$\begin{lgathered}\boxed{\begin{minipage}{25 em} \sf \displaystyle \sf\bullet\; a^{th} \; term\; of \;an \;AP\; \sf ( a_n ) = a + (n-1)d \\\\\\ \sf\bullet\; Sum\; of\; n \;terms \;of \;an \;AP \;( S_n ) = \dfrac{n}{2} \left(a + a_n\right) \\\\\\ \bullet\;\sf Sum \;of \;an \;AP\; having \;last \;term \;as\; l = \sf \dfrac{n}{2} \left(a + l\right) \end{minipage}}\end{lgathered}$