Question

8. Find four numbers in A.P. whose sum is
20 and the sum of whose squares is 120.
betwee​

Answer 1

$\large\bf\underline \blue {To \: \mathscr{f}ind:-}$

• we need to find four numbers in A.P whose sum is 20 and the sum of whose squares is 120.

$\huge\bf\underline \red{ \mathcal{S}olution:-}$

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

• Sum of 4 terms = 20
• sum of squares of 4 terms = 120

${ \blue{ \mathscr{ \underline{Let : - }}}}$

Let the four terms be( a - 3d ) , ( a - d) , (a + d) , (a + 3d)

Now,

$\underline{ \large \purple{ \mathscr{\dag\:A \bf{ccourding} \: to \: \mathscr {Q} \bf{uestion} ....}}}$

Sum of four terms of A.P is 20.

↛ ( a - 3d ) + ( a - d) + (a + d) + (a + 3d) = 20

↛ a - 3d + a - d + a + d + a + 3d = 20

↛ 4a = 20

↛ a = 20/4

↛ a = 5

Now,

sum of squares of 4 terms is 120

↛( a - 3d )² + ( a - d)² + (a + d)² + (a + 3d)² = 120

↛ a² + 9d² - 6ad + a² + d² - 2ad + a² + d² + 2ad + a² + 9d² + 6ad

↛ 4a² + 20d² = 120

• putting value of a

↛ 4 × 5² + 20d² = 120

↛ 4 × 25 + 20d² = 120

↛ 100 + 20d² = 120

↛20d² = 120 - 100

↛ 20d² = 20

↛d² = 1

↛ d = ±1

Hence,

• four terms are when d = 1 :-

➛ a - 3d = 5 - 3 × 1 = 5 - 3 = 2

➛ a - d = 5 -1 = 4

➛ a + d = 5 + 1 = 6

➛ a + 3d = 5 + 3 × 1 = 5 + 3 = 8

• four terms are when d = -1:-

➛ a - 3d = 5 - 3 × -1 = 5 + 3 = 8

➛ a - d = 5 - (-1 ) = 6

➛ a + d = 5 - 1 = 4

➛ a + 3d = 5 + 3 × -1 = 5 - 3 = 2

━━━━━━━━━━━━━━━━━━━━━━━━━

Answer 2

$\sf\large{\underline{Solution:-}}$

• Let the four term in A.P

be a-3d, a-d,a+d,a+3d.    ---- (1)

• Given that Sum of the terms = 20.

→ (a-3d) + (a-d) + (a+d) + (a+3d) = 20

→ 4a = 20

• →  a = 5  ---- (2)

• Given that sum of squares of the term = 120.

→ (a-3d)² + (a-d)² + (a+d)² + (a+3d)² = 120

→ (a² + 9d² - 6ad) + (a²+d²-2ab) + (a²+d²+2ad) + (a²+9d²+6ad) = 120

→ 4a² + 20d² = 120

• Substitute a = 5 from (2) .

→ 4(5)² + 20d² = 120

→ 100 + 20d² = 120

→ 20d² = 20

→ d = +1 (or) - 1.

• †Since AP cannot be negative† •

★ Substitute a = 5 and d = 1 in (1), we get

• a - 3d, a-d, a+d, a+3d

$\sf\large{\underline{Hence:-}}$

• The four term are → 2,4,6,8.