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# 2. the sum of the third and the seventh terms of an ap is 6 and their product is 8. find the sum of first sixteen terms of the ap.

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## The sum of the third and the seventh terms of an A.P is 6 and their product is 8 . Find the sum of first sixteen terms of the A.P

Click here👆to get an answer to your question ✍️ The sum of the third and the seventh terms of an A.P is 6 and their product is 8 . Find the sum of first sixteen terms of the A.P Question

## The sum of the third and the seventh terms of an A.P is 6 and their product is 8. Find the sum of first sixteen terms of the A.P

Let a and d be the first term and common difference of AP

Hard Open in App Solution Verified by Toppr nth term of AP a n ​ =a+(n−1)d ∴a 3 ​ =a+(3−1)d=a+2d a 7 ​ =a+(7−1)d=a+6d Given a 3 ​ +a 7 ​ =6 ∴(a+2d)+(a+6d)=6 ⇒2a+8d=6 ⇒a+4d=3....(1) Also given a 3 ​ ×a 7 ​ =8 ∴(a+2d)(a+6d)=8

⇒(3−4d+2d)(3−4d+6d)=8       [Using (1)]

⇒(3−2d)(3+2d)=8 ⇒9−4d 2 =8 ⇒4d 2 =1 ⇒d 2 = 4 1 ​ ⇒d=± 2 1 ​ When d= 2 1 ​ a=3−4d=3−4× 2 1 ​ =3−2=1 When d=− 2 1 ​ a=3−4d=3+4× 2 1 ​ =3+2=5 When a=1 & d= 2 1 ​ S 16 ​ = 2 16 ​ [2×1+(16−1)× 2 1 ​ ]=8(2+ 2 15 ​ )=4×19=76 When a=5 & d=− 2 1 ​ S 16 ​ = 2 16 ​ [2×5+(16−1)×(− 2 1 ​ )]=8(10− 2 15 ​ )=4×5=20

Thus, the sum of first 16 terms of the AP is 76 or 20.

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## The sum of the third and the seventh terms of an A.P is 6 and their product is 8 . Find the sum of first sixteen terms of the A.P.

The sum of the third and the seventh terms of an A.P is 6 and their product is 8 . Find the sum of first sixteen terms of the A.P. Arithmetic Progression

The sum of th... Question

The sum of the third and the seventh terms of an A.P is 6 and their product is 8. Find the sum of first sixteen terms of the A.P.

Open in App Solution We know that, = + (− 1) 3 = + (3 − 1) 3 = + 2 Similarly, 7 = + 6

Given that, 3 + 7 = 6

( + 2) + ( + 6) = 6 2 + 8 = 6 + 4 = 3 = 3 − 4 ()

Also, it is given that (3) × (7) = 8

( + 2) × (+ 6) = 8 From equation (), From equation (),  Suggest Corrections 36 SIMILAR QUESTIONS

Q. The sum of the third and the seventh terms of an

A P is 6 and their product is 8

. Find the sum of first sixteen terms of the

A P .

Q. The sum of the third and seventh term of an A.P. is

6

and their product is

8

. Find the sum of first sixteen terms of the A.P.

Q. The sum of the third and seventth term of an AP is 6 and their product is 8. Find the sum of the first 16 terms of the A.P.Q.

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

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## Ex 5.4, 2 (Optional)

Ex 5.4, 2 (Optional) The sum of the third and the seventh terms of an AP is 6 and their product is 8. Find the sum of first sixteen terms of the AP. We know that nth term of an AP is an = a + (n − 1)d Hence, 3rd term of AP = a3 = a + 2d and 7th term of AP = a7 = a + 6d Gi Check sibling questions

## Ex 5.4, 2 (Optional) - Chapter 5 Class 10 Arithmetic Progressions (Term 2)

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### Transcript

Ex 5.4, 2 (Optional) The sum of the third and the seventh terms of an AP is 6 and their product is 8. Find the sum of first sixteen terms of the AP. We know that nth term of an AP is an = a + (n − 1)d Hence, 3rd term of AP = a3 = a + 2d and 7th term of AP = a7 = a + 6d Given Sum of third & seventh terms is 6 a3 + a7 = 6 a + 2d + a + 6d = 6 2a + 8d = 6 2(a + 4d) = 6 a + 4d = 6/2 a + 4d = 3 a3 + a7 = 6 a + 2d + a + 6d = 6 2a + 8d = 6 2(a + 4d) = 6 a + 4d = 6/2 a + 4d = 3 Also, Product of the third and seventh terms is 8 a3 × a7 = 8 (a + 2d) (a + 6d) = 8 From (1) a + 4d = 3 a = 3 − 4d (3 − 4d + 2d) (3 − 4d + 6d) = 8 (3 − 2d) (3 + 2d) = 8 (3)2 − (2d)2 = 8 9 – 4d2 = 8 4d2 = 1 (2d)2 = (1)2 2d = ± 1 d = ± 𝟏/𝟐 Finding value of a For d = 𝟏/𝟐 a = 3 − 4d a = 3 − 4(1/2) a = 3 − 2 a = 1 For d = (−𝟏)/𝟐 a = 3 − 4d a = 3 − 4((−1)/2) a = 3 + 2 a = 5 For d = (−𝟏)/𝟐 a = 3 − 4d a = 3 − 4((−1)/2) a = 3 + 2 a = 5 Therefore, when a = 1, d = 𝟏/𝟐 And when a = 5, d = (−𝟏)/𝟐 Now, we need to find the Sum of first Sixteen Terms Sum of n terms of an AP is Sn = 𝒏/𝟐 [𝟐𝒂+(𝒏 −𝟏)𝒅] Taking a = 1 and d = 𝟏/𝟐 S16 = 16/2 [(2" × 1" )+(16 −1)(1/2)] = 8 [2+15/2] = 8 [(4 + 15)/2] = 8 × 19/2 = 76 Taking a = 5 and d = (−𝟏)/𝟐 S16 = 16/2 [(2" × " 5)+(16 −1)((−1)/2)] = 8 [10−15/2] = 8 [(20 − 15)/2] = 8 × 5/2 = 20 Hence, If a = 1 and d = 𝟏/𝟐 , the sum of first sixteen terms of the AP is 76 and If a = 5 and d = (−𝟏)/𝟐, the sum of first sixteen terms of the AP is 20.

Next: Ex 5.4, 3 (Optional) Important → 