Question

the sum pf the4th and 8th term of an AP is 54 and their product is 665.find the sum of 30 term of the AP (the terms of the AP are in ascending order )
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Answer 1

GIVEN :-

• The sum of the 4th and 8th term of an A.P is 54 and their product is 665.
• Terms of A.P are in ascending order. This means that Common difference is positive.

TO FIND :-

• Sum of first 30 terms of A.P.

SOLUTION :-

♦ For 4th term of A.P

$\\ \sf \: a_{4} = a + (4 - 1)d \\ \\ \sf \: a_{4} = a + 3d \: \: \: \: \: - - - (1)$

♦ For 8th term of A.P

$\\ \sf \: a_{8} = a + (8 - 1)d \\ \\ \sf \: a_{8} = a + 7d \: \: \: \: \: - - - (2)$

We know ,

$\\ \underline{\sf \: a_{4} + a_{8} = 54} \\ \\ \sf \: a + 3d + a + 7d = 54 \\ \\ \sf \: 2a + 10d = 54 \\ \\ \sf \: 2a = 54 - 10d \\ \\ \sf \: dividing \: whole \: equation \: by \: 2 \\ \\ \sf \: a = 27 - 5d \: \: \: \: \: \: \: \: \: - - - (3)$

Also ,

$\\ \underline{\sf \: \: a_{4} \times a_{8} = 665} \\ \\ \sf \: (a + 3d)(a + 7d) = 665 \\ \\ \sf \: a(a + 7d) + 3d(a + 7d) = 665 \\ \\ \sf \: {a}^{2} + 7ad + 3ad + 21 {d}^{2} = 665 \\ \\ \sf \: {a}^{2} + 10ad + 21 {d}^{2} = 665 \: \: \:$

Putting value of a from equation (3)

$\\ \sf \: (27 - 5 {d)}^{2} + 10(27 - 5d)(d) + 21 {d}^{2} = 665 \\ \\ \sf \: 729 - 270d + 25 {d}^{2} + 270d - 50 {d}^{2} + 21 {d}^{2} = 665 \\ \\ \sf \: 729 - 4 {d}^{2} = 665 \\ \\ \sf \: - 4 {d}^{2} = 665 - 729 \\ \\ \sf \: 4{d}^{2} = 64 \\ \\ \sf \: {d}^{2} = 16 \\ \\ \boxed{ \sf \: d = 4 , -4}$

But as we know that Common Difference is positive ,so , d = 4

Putting value of d in equation (3),

→ a = 27 - 5d

→ a = 27 - 5(4)

→ a = 27 - 20

→ a = 7

We have , a = 7 and d = 4

$\\ \boxed{\sf \: s_{n} = \dfrac{n}{2} \{2a + (n - 1)d \}} \\ \\ \implies\sf \: s_{30} = \dfrac{30}{2} \{2(7) + (30 - 1)(4) \} \\ \\ \implies\sf \: s_{30} = 15 \{14 + (29)(4) \} \\ \\ \implies\sf \: s_{30} = 15(14 + 116) \\ \\ \implies\sf \: s_{30} = 15(130) \\ \\ \implies\sf \: s_{30} = 1950$

Hence , sum of first 30 terms is 1950.

Answer 2

Answer:

1950

Step-by-step explanation:

According to the question, sum of 4th and 8th terms is 54

⇒ (a + 3d) + (a + 7d) = 54

⇒ 2a + 10d = 54

⇒ a + 5d = 27    

⇒ a = 27 - 5d    

Their product = 665

⇒ (a + 3d)(a + 7d) = 665

⇒ (27-5d + 3d)(27-5d + 7d) = 665

⇒ (27 - 2d)(27 + 2d) = 665

⇒ (27)² - (2d)² = 665

⇒ 729 - 4d² - 665

⇒ 64 = 4d²

⇒ 16 = d²

⇒ 4 = d         [d can't be -ve, as it must increase]

 thus, a = 27 - 5(4) = 7

 ∴ Sum of 1st 30 terms is

                 = (30/2) [2(7) + 29(4)]

                 = 15[14 + 116]

                 = 1950