Question

Joker and jaggu.... 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.​....Please don't answer ​

Answer 1

Answer:

Let a and d be the first term and common difference of A.P.

nth term of A.P., an = a + (n – 1) d

∴  a3 = a + (3 – 1) d = a + 2d

a7 = a + (7 – 1) d = a + 6d

Given, a3 + a7 = 6

∴ (a + 2d) + (a + 6d) = 6

⇒ 2a + 8d = 6

⇒ a + 4d = 3     ...(1)

Given, a3 × a7 = 8

∴ (a + 2d) + (a + 6d) = 8

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

⇒ (3 – 2d) (3 + 2d) = 8

⇒ 9 – 4d2 = 8

⇒ 4d2 = 1

a = 3 – 4d  = 3 – 2 = 1

When a = 1 and  

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

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Answer 2

Answer:

According to the conditions ;

a + 2d + a + 6d = 6

2a + 8d = 6

a + 4d = 3

a = 3 - 4d ----(1)

( a + 2d ) ( a + 6d ) = 8 ---(2)

substituting eq (1) in eq (2)

(3 - 4d + 2d )( 3 - 4d + 6d ) = 8

(3 - 2d)( 3 + 2d) = 8

9 - 4d^2 =8

4d^2 = 1

d^2 = 1 / 4

d = 1 / 2 ---(3)

substituting eq (3) in eq (1)

a = 3 - 4d

= 3 - 4 / 2

= 3 - 2

= 1

a = 1 and d = 1 / 2

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