Question

The sum of third and seventh terms of an arithmetic progression is 6 and rheir product is 8 .find?

Answer 1

Answer:

so using formula for n th term 

a_n=a+(n-1)dan=a+(n−1)d

a = first term

n = no. of terms 

d = common difference

So given

a + (3-1)d + a (7-1)d = 6

2a + 8d = 6

a + 4d = 3

a = 3 - 4d

(a + 2d)(a + 6d) = 8 

(3 - 2d)( 3 + 2d) = 8   {substituting a = 3 - 2d}

9 - 4d² = 8

d = +1/2 and - 1/2

so when d = +1/2 then a = 3 - 2 = 1 and when d = -1/2  a = 3 + 2 = 5

so using formula for 

Sn={n}{2}[2a+(n-1)d]

Sn=2n[2a+(n−1)d]

when d = +1/2

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

Sn=216[2∗1+(16−1)d]

Sn=76

Similarly when d = -1/2

Sn=20

Step-by-step explanation:

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