Question

2) The 6 th term of an A.P is 80 and its 15 th term is 16. Find the first 3
terms of the AP​

Answer 1

Answer :

The first three terms are  $\bf \frac{1040}{9},\frac{976}{9} ,\frac{912}{9}$

Step-by-step explanation :

Given,

• 6th term = 80
• 15th term = 16

To find,

• first 3 terms

Solution,

we know,

 nth term of A.P. is given by,

   $\boxed{\bf a_n=a+(n-1)d}$

where

a is the first term

d is the common difference

aₙ is the nth term

• 6th term = 80

      a₆ = a + (6 - 1)d

      80 = a + 5d --- [1]

• 15th term = 16

     a₁₅ = a + (15 - 1)d

     16 = a + 14d --- [2]

Subtract equation [1] from equation [2],

   16 - 80 = a + 14d - (a + 5d)

     - 64 = a + 14d - a - 5d

     - 64 = 9d

      d = -64/9

⇒ common difference = -64/9

Substitute d = -64/9 in equation [1]

     80 = a + 5d

     80 = a + 5(-64/9)

     80 = a - (320/9)

     a = 80 + (320/9)

     a = (720+320)/9

     a = 1040/9

first term,

a = 1040/9

second term,

a₂ = a + d

a₂ = 1040/9 + (-64/9)

a₂ = (1040-64)/9

a₂ = 976/9

third term,

a₃ = a₂ + d

a₃ = 976/9 + (-64/9)

a₃ = (976-64)/9

a₃ = 912/9

Answer 2

Given:-

$\bigstar$let a is the first term and d is common difference of the AP.

$\bigstar$6th term of an AP is 80

To Prove:-

$\bigstar$The first 3 terms of the AP

Solution:-

$\\ \sf \: a + (6 - 1) = a + 5d = 80 \longrightarrow(1)$

$\bigstar$and its 15th term is 16.

$\\ \sf \: a + (15 - 1) = a + 14d = 16 \longrightarrow(2)$

$\bigstar$from equations (1) and (2) we get,

$\\ \sf \: a + 14d - a - 5d = 16 - 80 = - 64$

$\\ \sf \: 9d \: = - 64$

$\\ \sf \: d = \frac{ - 64}{9}$

$\\ \sf \: a = 80 - 5d = 80 - 5 ( \frac{ - 64}{9}) = 80 + \frac{320}{9}$

$\\ \sf \: \frac{1040}{9}$

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