Question

The sum of an AP is 21 and product of the first and third term exceeds the second term by 6. Find the three terms​

Answer 1

Step-by-step explanation:

The sum of three terms of an A.P. is 21 and the product of the first and the third terms exceeds the second term by 6, find three terms. dd=±√36=±6.mark me as brainliest

Answer 2

ɢɪᴠᴇɴ:-

• The sum of an A.P is 21 and product of the first and third term exceeds the second term by 6

ᴛᴏ ғɪɴᴅ:-

• The next three terms of the AP.

ᴇxᴘʟᴀɴᴀᴛɪᴏɴ:-

We know that formula of an A.P;

$\boxed{\bf{a_{n}=a+(n-1)d}}}}}$

Here,

• a is the first term.

• d is the common difference.

• n is the term of an A.P.

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.

$\begin{lgathered}\mapsto\sf{(a - d) + (a) + (a + d) = 21}\\\\\mapsto\sf{a - d + a + a + d = 21}\\\\\mapsto\sf{a=\cancel{\dfrac{21}{3} }}\\\\\mapsto\sf{\pink{a=7}}\end{lgathered}$

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

$\green{ \underline{ \boxed{ \sf{d=6}}}}$

Therefore, the next three terms of the AP be......

$\begin{gathered}\\\\\bullet\sf{(a-d)=(7-6)=\boxed{1}}}}\\\\\bullet\sf{a=7=\boxed{16}}}\\\\\bullet\sf{a+d=7+6=\boxed{13}}}\end{gathered}$

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