Question

the sum of the 3rd and 7th termnof an ap is 6 and their product is 8 find the sum of first 16 term of an ap​

Answer 1

Answer:

Step-by-step explanation:

Let the first term be a and common difference be d

nth term = a+(n-1)d

Given Third Term + Seventh term = (a+2d)+(a+6d) = 6 ==> a+4d = 3

hence, a= 3-4d

Third Term * Seventh term = (a+2d)*(a+6d) = 8

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

i.e. 9-4d^2 = 8==> d^2 = (9-8)/4 = 0.25==> d = 0.5 or -0.5

Now to check which is correct d...

Substitute and find

Case (a): d= 0.5

a+4d = 3==> a=3-4d = 3-4(0.5)=1

3rd term = a+2d= 1+2*0.5 = 2

7th term = a+6d= 1+6*0.5 = 4

Sum = 6 and Product = 8

Case (b): d= -0.5

a+4d = 3==> a=3-4d = 3-4(-0.5) = 3+2 = 5

3rd term = a+2d= 5+2*(-0.5) = 4

7th term = a+6d= 5+6*(-0.5) = 2

Sum = 6 and Product = 8

Since both are matching, we will go with bothvalues

Sum of first 16 terms = n*(2a+(n-1)d)/2 = 16*(2a+15d)/2

= 8*(2a+15d)

Case (a): d= 0.5

Sum = 8*(2*1+15*0.5)=76

Case (b): d= 0.5

Sum = 8*(2*5+15*(-0.5))=20..

Hope it helps you

Answer 2

The 3rd term of an AP is a+2d and the 7th term is a+6d. So we have

the sum of 3rd term and 7th term = a+2d+a+6d = 6, or

2a+8d = 6, or

a + 4d = 3 …(1)

Their product is (a+2d)*(a+6d) = 8 …(2)

From (1), a = 3–4d. Put that in (2) to get

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

(3–4d+2d)*(3–4d+6d) = 8 or

(3–2d)*(3+2d) = 8

9–4d² = 8

4d² = 1

d² = 1/4

d = 1/2.

From (1), a = 3–4d = 3–2 = 1.

The sum of the first 16 terms

S16 = (n/2)[2a+(n-1)d]

= (16/2)[2*1 + (16–1)*1/2]

= 8[2+15/2]

= 8 *19/2

= 76.