Question

the sum of first three terms of an Ap is 48 if the producd of first and second term execds 4 times the third term by 12 find ap
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Answer 1

Answer:

The first three terms of the AP is 7,16,25.

Step-by-step explanation:

Given :-

• The sum of first three terms of an AP is 48.
• The product of first and second term exceeds 4 times the third term by 12.

To find :-

• The AP.

Solution :-

Let the first three terms of the AP be (a-d), a and (a+d).

• a = first term
• d = common difference

According to the 1st condition ,

• The sum of first three terms of an AP is 48.

(a-d)+a+(a+d) = 48

→ a-d+a+a+d = 48

→ 3a = 48

→ a = 48/3

→ a = 16

According to the 2nd condition,

• The product of first and second term exceeds 4 times the third term by 12.

(a-d)×a = 4(a+d)+12

→ (16-d)×16= 4(16+d)+12

→ 256-16d= 64+4d+12

→ -16d-4d = 64+12-256

→ -20d = -180

→ 20d = 180

→ d = 180/20

→ d = 90

Therefore,

• (a-d) = (16-9) = 7
• a = 16
• (a+d) = (16+9) = 25

Hence the first three terms of the AP are 7,16,25.

Answer 2

Solution :

Let the first three terms of the AP be (a - d), a, (a + d).

____________________

Then, According to given condition (1) we get:

(a - d) + a + (a + d) = 48

3a = 48

a = 48/3

a = 16

____________________

Now, According to condition (2) we get:

(a - d) × a = 4 (a + d) + 12

(16 - d) × 16 = 4 (16 + d) + 12

256 - 16d = 64 + 4d + 12

16d + 4d = 256 - 76

20d = 180

d = 180/20

d = 9

......----------------------------------......

By putting the value of a and d in (a - d) and (a + d) we get,

• (a - d) = 16 - 9 = 7
• (a + d) = 16 + 9 = 25

Therefore, the first three terms of the AP are 7, 6, 25.