Question

The sum of three consecutive terms of an Ap is 24 and their product is 224
*write the 1st 3terms of this sequence?
*write the algebraic form of this Ap?
*Is 2020 a term of this sequence?
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Answer 1

Given : -

Sum of 3 consecutive terms of an AP is 24 .

Product of 3 consecutive terms is 224 .

Required to find : -

• 1st 3 terms of the AP ?

• Algebraic form of this AP ?

• Is 2020 a term of this AP ?

Solution : -

Sum of 3 consecutive terms of an AP is 24 .

Product of 3 consecutive terms is 224 .

We need to find ;

• 1st 3 terms of the AP ?

• Algebraic form of this AP ?

• Is 2020 a term of this AP ?

So,

Now,

Let's consider the first statement ;

Sum of 3 consecutive terms of an AP is 24 .

Let the 1st term be a - d

2nd term be a

3rd term be a + d

According to problem ;

➠ a - d + a + a + d = 24

➠ 3a = 24

➠ a = 24/3

➠ a = 8

• Value of a = 8

Similarly,

➠ ( a - d ) ( a ) ( a + d ) = 224

➠ ( value of a = 8 ) = 224

➠ ( 8 - d ) ( 8 ) ( 8 + d ) = 224

➠ ( 8 - d ) ( 8 + d ) = 224/8

➠ ( 8 - d ) ( 8 + d ) = 28

➠ ( 8 )² - ( d )² = 28

➠ 64 - d² = 28

➠ 64 - 28 = d²

➠ 36 = d²

➠ d² = 36

➠ d = √36

➠ d = ± 6

➠ d = + 6 cm or - 6 cm

➠ d = 6 cm

Here we need to apply a bit of logic , i.e. the first term + common difference > 8 that is why + 6 is the common difference . Moreover, the sum of three consecutive terms is also in positive.

Hence,

• First term ( a ) = 8

• Common difference ( d ) = 6

$\mathtt{1st \: term = a = \boxed{ 8}} \\ \\ \mathtt{2nd \: term = a + d = 8 + 6 = \boxed{14}} \\ \\ \mathtt{3rd \: term = a + 2d = 8 + 2(6) = 8 + 12 = \boxed{ 20}}$

Hence,

• 1st term = 8 , 2nd term = 14 & 3rd term = 20

Now,

Let's find the algebraic form of the AP .

Using the formula ;

$\boxed{\sf{ {a}_{nth} = a + ( n - 1 ) d }}$

Here,

• a = first term

• d = common difference

• n = term number

$\sf{ {a}_{nth} = 8 + ( n - 1 ) 6 } \\ \\ \sf{ a_{nth} = 8 + 6n - 6 } \\ \\ \sf{ a_{nth} = 2 + 6n }$

Hence,

• Algebraic form of the AP = 2 + 6n

Now,

Let's find out whether 2020 is the term this sequence or not

Using the formula ;

$\boxed{\sf{ {a}_{nth} = a + ( n - 1 ) d }}$

Here, nth term = 2020

➠ 2020 = 8 + ( n - 1 ) 6

➠ 2020 = 8 + 6n - 6

➠ 2020 = 2 + 6n

➠ 2020 - 2 = 6n

➠ 2018 = 6n

➠ 6n = 2018

➠ n = 2018/6

➠ n = 336.33 . . . . .

Since,

Term number should be a natural number .

Hence,

• 2020 is not the term of this AP

Answer 2

Step-by-step explanation:

a) Assume the three consecutive terms be (a-d), (a), (a+d).

As per given condition,

→ a - d + a + a + d = 24

→ 3a = 24

→ a = 8

→ (a - d)(a)(a + d) = 224

→ (8 - d)(8)8 + d) = 224 [From above]

→ (8 - d)(8 + d) = 224/8

Used identity: (a - b)(a + b) = a² - b²

→ (8)² - (d)² = 28

→ 64 - d² = 28

→ 64 - 28 = d²

→ d² = 36

→ d = 6

First term is 8 and common difference is 6.

• a - d = 8 - 6 = 2
• a = 8
• a + d = 8 + 6 = 14

Now,

b) an = a + (n - 1)d

an = 8 + (n - 1)6

an = 8 + 6n - 6

an = 2 + 6n

c) 2020 = 8 + (n - 1)6

2020 - 8 = (n - 1)6

2012 = (n - 1)6

335.33 = (n - 1)

336.33 = n