Question

The sum of first three terms of an AP is 33. The product of first and last terms is 29 more than the second term. Find the A.P.

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Answer 1

GiveN :

• Sum of first three term of An AP is 33.
• Product of first and last terms is 29 more than the second term.

To FinD :

Find the AP.

SolutioN :

Let,

• First term " a "
• Common difference " d "
• Sum of nth term " Sn "

A/Q,

Case 1.

• Sum of first three term of An AP is 33. Product of first and last terms is 29.

Let,

• a , a + d and a + 2d.

$\tt : \implies a + a + d + a + 2d = 33.$

$\tt : \implies 3a + 3d= 33.$

$\tt : \implies 3(a + d) = 33.$

$\tt : \implies a + d = 11.$

$\tt \dagger \: \: \: \: \: d = 11 - a\: \: \: \: \: - (1)$

Now,

• Product of first and last terms is 29 more than the second term.
• a , a + 2d , a + d.

$\tt : \implies (a)(a + 2d) = a + d$

$\tt : \implies (a)(a + 2(11 - a)) = a + 11 - d$

$\tt : \implies (a)(a + 22 - 2a) = 11 + 29.$

$\tt : \implies a(22 - a) = 40.$

$\tt : \implies { - a}^{2} + 22 a= 40.$

$\tt : \implies { - a}^{2} + 22a - 40 = 0.$

$\tt : \implies { a}^{2} - 22a + 40 = 0.$

$\tt : \implies { a}^{2} - 20a - 2a+ 40 = 0.$

$\tt : \implies a(a - 20) - 2(a - 20) = 0.$

$\tt : \implies (a - 20) (a - 2) = 0.$

$\tt : \implies a = 20 \: \: \: or \: \: \: 2.$

Now,Form Equation 1.

$\tt : \implies d = 11 - a.$

$\tt : \implies d = 11 - 20$

$\tt : \implies d = - 9.$

Or,

$\tt : \implies d = 11 - a$

$\tt : \implies 11 - 2$

$\tt : \implies d = 9.$

NoW AP

★ Case 1

• When a = 20 and d = -9.

→ a = 20.

→ a + d = 20 + ( - 9 ) = 11.

→ a + 2d = 20 + 2( -9 ) = 4.

AP → 20 , 11 , 2.

★ Case 2.

• When, a = 2 and d = 9.

→ a = 2.

→ a + d = 2 + 9 = 11.

→ a + 2d = 2 + 2( 9 ) = 20.

AP → 2 , 11 , 20.

Answer 2

Given :

• Sum of first three terms of A.P = 33.
• Product of first and last term = 29 more than second term.

To Find :

• Arithmetic progression (A.P)

Solution :

Let the first, second and third terms of the A.P be (a-d), a and (a+d).

Then, according to the question

★Case I ;

→(a-d) + a + (a+d) = 33

→a + a+ a +d - d = 33

→3a = 33

→a = 33/3 = 11.

And,

★Case II ;

$→(a - d)(a + d) = 29 + a \\ \\ →{a}^{2} - {d}^{2} = 29 + a \\ \\ → {11}^{2} - {d}^{2} = 29 + 11 \\ \\ →121 - {d}^{2} = 40 \\ \\→ 121 - 40 = {d}^{2} \\ \\ →81 = {d}^{2} \\ \\ → \sqrt{81} = 9 = d$

Thus, the A.P is (11-9),11,(11+9) i.e 2, 11, 20... so on.