Question

find the three consecutive term​

Answer 1

➤Question :-

$\rm \red{Find \: three \: consecutive \: terms \: in \: AP} \: \\ \rm \green{whose \: sum \: is \: 9 \: and \: product \: of \: their }\\ \rm \: cube \: is \: 3375.$

➤Answer :-

→ Three consecutive terms are → 1 , 3 and 5 .

➤Solution :-

$\rm Let \: three \: consecutive \: terms \: are \: \\ \to \rm (x - d) ,( x) , (x+d )\: \\ \\ \rm \pink{ According \: to \: the \: question} \\ \\ \mapsto \rm \: sum \: of \: these \: terms \: = 9 \\ \\ \mapsto \rm \: ( x - d )+ (x )+( x + d )= 9 \\ \\ \mapsto \rm 3x = 9 \\ \\ \mapsto \boxed{\rm x = 3} \\ \\ \sf \: \: \: \orange{and \: also \:} given \: that \\ \\ \to \rm product \: of \: their \: cube \: = 3375 \\ \\ \to \rm {x}^{3} \: {(x - d)}^{3} {(x + d)}^{3} = 3375 \\ \\ \because \rm \: \: \: x = 3 \\ \\ \to \rm {3}^{3} {(3 - d)}^{3} {(3 + d)}^{3} = 3375 \\ \\ \to \rm { \{(3 - d)(3 + d) \}}^{3} = \frac{3375}{27} \\ \\ \: \: \: \: \: \: \: \boxed{\because \rm {(a + b)(a - b)} = {a}^{2} - {b}^{2} } \\ \\ \therefore \\ \to \rm { \{ {3}^{2} - {d}^{2} \}}^{3} = 125 \\ \\ \sf taking \: \sqrt[3]{} \: on \: both \: sides \\ \\ \to \rm \sqrt[3]{ {(9 - {d}^{2} )}^{3} } = \sqrt[3]{125} \\ \\ \to \rm 9 - {d}^{2} = 5 \\ \\ \to \rm {d}^{2} = 4 \\ \\ \to \rm \: d = + 2 \: \: or \: - 2 \\ \\ \sf \green{ difference \: between \: terms \: can }\: not \: negative \\ \\ \to \boxed{ \rm d = 2} \\ \\ \sf \red{ therefore \: required \: terms \: are \: } \\ \\ \to \star \boxed{ \rm first \: term \: = x - d = 3 - 2 = 1} \\ \\ \to \star \boxed{ \rm second \: term = x = 3 }\\ \\ \to \star \boxed{ \rm third \: term = x + d = 3 + 2 = 5}$

Verification :-

Sum of terms = 9

→ 1 + 3 + 5 = 9

→ 9 = 9

and ,product of their cube = 3375

→ (1)³ × (3)³ ×(5)³ = 3375

→ 27 × 125 = 3375

→ 3375 = 3375

Hence verified .