# 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. - Mathematics

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.

#### Solution

We know that,

an = a + (− 1) d

a3 = a + (3 − 1) d

a3 = a + 2d

Similarly, a7 = a + 6d

Given that, a3 + a7 = 6

(a + 2d) + (a + 6d) = 6

2a + 8d = 6

a + 4d = 3

a = 3 − 4d (i)

Also, it is given that (a3) × (a7) = 8

(a + 2d) × (+ 6d) = 8

From equation (i),

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

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

9 - 4d2 = 8

4d2 = 9 - 8 =1

d2 =  1/4

d = ± 1/2

d = 1/2 or 1/2

From the equation (i)

(When d os 1/2)

a = 3- 4d

a = 3 -4(1/4)

= 3 - 2 = 1

(When d is - 1/2)

a = 3 - 4(-1/2)

a = 3+2 = 5

S_n = n/2 [2a(n-1)s]

(when a is  1 and d is 1/2 )

S_16 = 16/2[2(1)+(16+1)(1/2)]

= 8[2+15/2]

= 4(19) =76

(When a is 5 and d is -1/2)

S_16=16/2[2(5)+(16-1)(-1/2)]

= 8[10+(15)(-1/2)]

= 8(5/2)

= 20

Concept: Sum of First n Terms of an AP
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#### APPEARS IN

NCERT Class 10 Maths
Chapter 5 Arithmetic Progressions
Exercise 5.4 | Q 2 | Page 115