Question

Sum of first six terms of an Arithmetic Progression is 6, product of second and fifth
term is -80. Write the terms of Arithmetic Progression​

Answer 1

Arithmetic Progression :-

Note:

• A sequence is said to be in AP (Arithmetic Progression), if the difference between its consecutive terms are equal.

• The nth term of an AP is given as ;

T(n) = a + (n-1)•d , where a is the first term and d is the common difference.

• The common difference of an AP is given as ;

d = T(n) - T(n-1)

• If the number of terms in an AP is n ( where n is odd ) ,then there will be a single middle term.

Also, [(n+1)/2]th term will be its middle term.

• If the number of terms in an AP is n ( where n is even ) ,then there will be two middle terms.

Also, (n/2)th and (n/2 + 1)th terms will be its middle terms.

• The sum up to nth terms of an AP is given as ;

S(n) = (n/2)•[2a + (n-1)•d] where a is the first term and d is the common difference.

• The nth term of an AP is also given as ;

T(n) = S(n) - S(n-1)

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Solution :-

→ Let First Term of AP = a

→ common Difference = d

So,

→ Sn = (n/2)•[2a + (n-1)•d]

→ S(6) = (6/2) [ 2a + (6-1)d]

→ 6 = 3[2a + 5d]

→ 2 = 2a + 5d

→ (2 - 5d) = 2a

→ a = (2 - 5d)/2 -------------------- Equation (1).

________________________

Now , we have given That, product of second and fifth

product of second and fifthterm is -80.

So,

→ Second Term = a + d

→ 5th Term = a + (5-1)d = a + 4d

A/q,

→ (a + 4d)(a + d) = (-80)

Putting value of Equation (1) now we get,

→ [ {(2 - 5d)/2} + 4d ] [{(2 - 5d)/2} + d ] = (-80)

→ [ (2 - 5d + 8d) /2 ] [ (2 - 5d + 2d) / 2 ] = (-80)

→ (2 + 3d)(2 - 3d) = (-80)*4

→ (2)² - (3d)² = -320

→ 9d² = 4 + 320

→ 9d² = 324

→ d² = 36

→ d = ± 6

Putting This Value in Equation (1) , we get,

→ a = (2 - 5d)/2 = (2 - 5*6) / 2 = (-28)/2 = (-14)

Or,

→ a = (2 - 5d)/2 = (2 +5*6) / 2 = (32/2) = 16.

__________________

So, AP will be :-

when a = 16, d = (-6)

→ 16, 10 , 4 , (-2) , (-8) , (-14) .

when a = (-14) , d = 6

→ (-14) , (-8) , (-2) , 4, 10, 16 .

___________________

[ Nice Question. ]

Answer 2

Answer:

AP : 16, 10, 4, - 2, - 8, - 14....

Or,

AP : - 14, - 8, - 2, 4, 10, 16...

Step-by-step explanation:

Given that :

• The sum of first six terms of an AP is 6.
• Product of second and fifth term is - 80.

To find :

• The terms of Arithmetic Progression.

Solution :

Let the terms of the given AP be a, a + d, a + 2d, a + 3d,.... a + nd.

CASE I :

The sum of first six terms of the AP is 6.

We know that,

$\sf \implies S_n = \frac{n}{2} [ 2a + (n-1)d] \\\\\implies \sf S_6 = \frac{6}{2} [2a + (6-1)d]\\\\\implies \sf 6 = 3 (2a + 5d)\\\\\implies \sf 2a + 5d = 2 \\\\\implies \sf a = \frac{2 - 5d}{2} \: \: \dots (i)$

CASE II :

Product of second and fifth term is - 80.

$\sf \implies t_2 \times t_5 = - 80\\\\\implies \sf (a + d)(a + 4d) = - 80$

[ Using equation (i) ]

$\implies \sf \left(\frac{2 - 5d}{2} + d\right) \left(\frac{2 - 5d}{2} + 4d\right) = - 80\\\\\implies \sf \left(\frac{2 - 5d + 2d}{2} \right)\left(\frac{2 - 5d + 8d}{2} \right) = - 80 \\\\\implies \sf \left(\frac{2 - 3d}{2}\right) \left(\frac{2 +3d}{2} \right) = - 80 \\\\\implies \sf (2 - 3d)(2+3d) = - 80 * 4\\\\\implies \sf (2)^2 - (3d)^2 = - 320 \\\\\implies \sf 4 - 9d^2 = - 320\\\\\implies \sf - 9d^2 = - 320 - 4 \\\\\implies \sf - 9d^2 = - 324 \\\\\implies \sf d^2 = \frac{-324}{-9} \\\\\implies \sf d^2 = 36 \\\\\implies \sf d = \pm 6$

So, we get two values of d.

For d = - 6,

$\implies \sf a = \frac{2 - 5d}{2} \\\\\implies \sf a = \frac{2-5(-6)}{2} \\\\\implies \sf a = \frac{2+30}{2} \\\\\implies \sf a = \frac{32}{2} \\\\\implies \sf a = 16$

For d = 6,

$\implies \sf a = \frac{2 - 5d}{2} \\\\\implies \sf a = \frac{2-5(6)}{2} \\\\\implies \sf a = \frac{2-30}{2} \\\\\implies \sf a = \frac{-28}{2} \\\\\implies \sf a = - 14$

Arithmetic Progression with first case :

When, a = 16 and d = - 6 ;

⇒ AP : 16, 16 - 6, 16 - 12....

⇒ AP : 16, 10, 4, - 2, - 8, - 14....

Arithmetic Progression with second case:

When, a = - 14 and d = 6 ;

⇒ AP : - 14, - 14 + 6, - 14 + 12....

⇒ AP : - 14, - 8, - 2, 4, 10, 16...