Question

Sum of for consecutive numbers in ap is 20 and the ratio of product of the First and last terms to the product of the two middle terms is 2:3

Answer 1

$\large\bf\underline{Correct\: Question:-}$

sum of four consecutive numbers in ap is 20 and the ratio of product of the First and last terms to the product of the two middle terms is 2:3. Find the numbers.

━━━━━━━━━━━━━━━━━━━

$\large\bf\underline{Given:-}$

• Sum of four consecutive numbers in AP is 20
• The ratio of product of the First and last terms to the product of the two middle terms is 2:3.

$\large\bf\underline {To \: find:-}$

• Numbers

$\huge\bf\underline{Solution:-}$

Let the four consecutive numbers be :-

• ≫ (a- 3d) ,(a -d), (a+d) ,(a + 3d)

$\underbrace{ \blacktriangleright \: \large \bf \: According \: to \: question}$

Sum of for consecutive numbers in ap is 20.

$\small \rightarrowtail \rm \: (a - 3d) + (a - d) + (a + d) + (a + 3d) = 20 \\ \\ \small \rightarrowtail \rm \:a - \cancel{ 3d }+ a \cancel{ - d }+ a + \cancel{ d }+ a + \cancel{ 3d} = 20 \\ \\ \small \rightarrowtail \rm \:4a = 20 \\ \\ \small \rightarrowtail \rm \:a = \cancel\dfrac{20}{4} \\ \\ \small \rightarrowtail \bf \boxed{ \bf \: a = 5}$

And,

the ratio of product of the First and last terms to the product of the two middle terms is 2:3

$\rightarrowtail \rm \: \frac{(a - 3d)(a + 3d)}{(a - d)(a + d)} = \frac{2}{3} \\ \\ \rightarrowtail \rm \: \frac{ {a}^{2} - {9d}^{2} }{ {a}^{2} - {d}^{2} } = \frac{2}{3} \\ \\ \rightarrowtail \rm \:3( {a {}^{2} - {9d}^{2}}) = 2( {a}^{2} - {d}^{2} ) \\ \\ \rightarrowtail \rm \:3 {a}^{2} - 27 {d}^{2} = 2 {a}^{2} - {2d}^{2} \\ \\ \rightarrowtail \rm \:3 {a}^{2} - 2 {a}^{2} = - 2 {d}^{2} + 27 {d}^{2} \\ \\ \rightarrowtail \rm \: {a}^{2} = 25 {d}^{2} \\ \rm \dag \: putting \: value \: of \: a = 5 \: we \: get \\ \\ \rightarrowtail \rm \:5 {}^{2} = 25 {d}^{2} \\ \\ \rightarrowtail \rm \:25 = 25 {d}^{2} \\ \\ \rightarrowtail \rm \: {d}^{2} = \cancel\dfrac{25}{25} \\ \\ \rightarrowtail \rm \boxed{\bf\:d = 1}$

So,

Four consecutive numbers are :-

»» a - 3d = 5 - 3× 1 = 2

»» a -d = 5 - 1 = 4

»» a+d = 5 + 1 = 6

»» a + 3d = 5 +3 × 1 = 8

So,

≫Four consecutive numbers are: 2,4,6,8

Answer 2

Let , the four consecutive terms of AP be (a - 3d) , (a - d) , (a + d) , and (a + 3d)

First Condition :

The sum of four consecutive terms of AP is 20

$\sf \mapsto a - 3d + a - d + a + d + a + 3d = 20 \\ \\ \sf \mapsto </p><p>4a = 20 \\ \\ \sf \mapsto</p><p>a = \frac{20}{4} \\ \\ \sf \mapsto</p><p>a = 5$

2nd Condition :

The ratio of product of the first and last terms to the product of the two middle terms is 2 : 3

$\sf \mapsto \frac{(a - 3d)(a + 3d)}{(a - d)(a + d)} = \frac{2}{3} \\ \\ \sf \mapsto\ \frac{{(a)}^{2} - {(3d)}^{2} }{ {(a)}^{2} - {(d)}^{2} } = \frac{2}{3} \\ \\ \sf \mapsto \frac{ {(5)}^{2} - {(3d)}^{2} }{ {(5)}^{2} - {(d)}^{2} } = \frac{2}{3} \\ \\ \sf \mapsto75 - 27 {(d)}^{2} = 50 - 2 {d}^{2} \\ \\ \sf \mapsto 25{(d)}^{2} = 25 \\ \\ \sf \mapsto {(d)}^{2} = 1 \\ \\ \sf \mapsto d = 1$

$\therefore \sf \underline{The \: four \: consecutive \: terms \: of \: AP \: are \: 2 \: , \: 4 \: , \: 6 \: , \: and \: 8}$