Question

The sum of the first three terms of om AP is 33
is the product of the first term and third term exceeds the second term by 29. find the AP​

Answer 1

Answer:

• Terms of the AP are 2, 11, 20.
• Or, 20, 11, 2.

Step-by-step explanation:

Given that:

• The sum of the first three terms of an AP is 33.
• The product of the first term and third term exceeds the second term by 29.

To Find:

• Terms of the AP.

Let us assume:

• First term = a - d
• Second term = a
• Third term = a + d

Sum of the first three terms:

⟿ (a - d) + a + (a + d) = 33

⟿ a - d + a + a + d = 33

⟿ 3a = 33

⟿ a = 33/3

⟿ a = 11

Product of the first term and third term:

⟿ (a - d) (a + d) = a + 29

Substituting the value of a.

⟿ (11 - d) (11 + d) = 11 + 29

⟿ 11² - d² = 40

⟿ 121 - d² = 40

⟿ d² = 121 - 40

⟿ d² = 81

⟿ d = √81

⟿ d = ± 9

Terms of the AP:

When d = + 9

• First term = a - d = 11 - 9 = 2
• Second term = a = 11
• Third term = a + d = 11 + 9 = 20

When d = - 9

• First term = a - d = 11 - (- 9) = 20
• Second term = a = 11
• Third term = a + d = 11 + (- 9) = 2

Hence,

• Terms of the AP are 2, 11, 20.
• Or, 20, 11, 2.

Answer 2

Step-by-step explanation:

Let

the terms of AP are-(a-d),a,(a+d)

ATQ

$\\ \sf \longmapsto (a-d)+a+(a+d)=33$

$\\ \sf \longmapsto a-d+a+a+d=33$

$\\ \sf \longmapsto 3a=33$

$\\ \sf \longmapsto a=\dfrac{33}{3}$

$\\ \bf \longmapsto a=11$

Again

$\\ \sf \longmapsto (a-d)(a+d)=a+29$

$\\ \sf \longmapsto a^2-d^2=a+29$

• Substituting a's value

$\\ \sf \longmapsto 11^2-d^2=11+29$

$\\ \sf \longmapsto 121-d^2=40$

$\\ \sf \longmapsto 121-40=d^2$

$\\ \sf \longmapsto d^2=81$

$\\ \sf \longmapsto d=\sqrt{81}$

$\\ \bf \longmapsto d=\underline{+}9$

if d=+9

$\\ \sf \longmapsto a-d=11-9=2$

$\\ \sf \longmapsto a=11$

$\\ \sf \longmapsto a+d=11+9=20$

Thus 2,11,20 are in AP.

if d=-9

$\\ \sf \longmapsto a-d=11-(-9)=11+9=20$

$\\ \sf \longmapsto a=11$

$\\ \sf \longmapsto a+d=11+(-9)=11-9=2$

Thus 20,11,2 are in AP.