Question

The sum of three terms of an



a.P is 21 and the product of first and third terms exceeds the second term by 6, find the three terms

Answer 1

Step-by-step explanation:

let the three terms of ap are

(a - d) , (a) , (a + d)

As we have given condition, The sum of three terms of an a.P is 21.

→ (a - d) + (a) + (a + d) = 21

→ a - d + a + a + d = 21

→ 3a = 21

→ a = 21 / 3

$\underline{\textbf{\large{a = 7}}}$

Now, As we have given the product of first and third terms exceeds the second term by 6

→ (a - d) (a + d) = (a) + 6

→ a² - d² = a + 6

put the value of a

→ (7)² - d² = (7) + 6

→ 49 - d² = 13

→ -d² = 13 - 49

→ -d² = -36

→ d² = 36

$\underline{\textbf{\large{ d = 6 }}}$

therefor the three terms are

$\boxed{\underline{\textbf{\large{(a - d) = (7 - 6 ) = 1}}}}$

$\boxed{\underline{\textbf{\large{(a)= 7}}}}$

$\boxed{\underline{\textbf{\large{(a + d) = ( 7 + 6 ) = 13}}}}$

Answer 2

• Let three terms be (a - d), (a), (a + d)

» The sum of three terms of an A.P is 21

=> a - d + a + a + d = 21

=> 3a = 21

=> a = 21/3

=> a = 7 ______ (eq 1)

____________________________

» The product of first and third terms exceeds the second term by 6.

First term is (a - d)

Second term is (a)

Third term is (a + d)

=> (a - d) (a + d) = (a) + 6

We know that

(a - b) (a + b) = a² - b²

=> a² - d² = a + 6

=> (7)² - d² = 7 + 6

=> 49 - d² = 13

=> - d² = - 36

=> d = √36

=> d = ± 6 ________ (eq 2)

______________________________

Now.. take a = 7 and d = + 6

• First term = (a - d)

=> 7 - 6

=> 1

• Second term = (a)

=> 7

• Third term = (a + d)

=> 7 + 6

=> 13

» Take a = 7 and d = - 6

• First term = (a - d)

=> 7 - (-6) = 7 + 6

=> 13

• Second term = (a)

=> 7

• Third term = (a + d)

=> 7 - 6

=> 1

_____________________________

1, 7 and 13 are the three terms.

___________ [ ANSWER ]

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