Question

the sum of the third and seventh term of an AP is 6 and their product is 8 find the first term and the common difference​

Answer 1

Answer:

Step-by-step explanation:

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 d = +1/2

S_n=\frac{16}{2}[2*1+(16-1)d]

S_n=76

=>Similarly when d = -1/2

S_n=20

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

Hope it helps you.......!!!!!

Answer 2

Answer:

Step-by-step explanation: