Question

answeerrrr fast plz​

Answer 1

Answer:

68

Step-by-step explanation:

given

sum of third and seventh term is 6

in a.p the general term is

an = a + (n - 1)d

third term is

》a3 = a + (3 -1)d

》a3 = a + 2d

similarly

seventh term

》a7 = a + 6d

》》a3 + a7 = (a + 2d) + (a + 6d) = 2a + 8d = 6

.......(given)

》2a + 6d = 6

》a + 4d = 3 .....eq 1

also given product is 8

a3*a7 = I

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

》a^2 + 12d^2 + 8ad = 8 ....eq 2

by squaring (eq 1) and subtracting with (eq 2)

we get

(a^2 + 12d^2 + 8ad) - (a +4d)^2 = 8 - 3^2

》(a^2 + 12d^2 + 8ad) - (a^2 + 16d^ 2 + 8ad) = 8 - 9

》-4d^2 = -1

》d^2 = 1/4

》d = 1/2

substitute this in eq 1

a + 4d = 3

》a + 4*1/2 = 3

》a = 1

sum of first 16 term = sn = n/2 * (a + (n - 1)d)

》s16 = 16/2 * (1 + (16 - 1) * 1/2)

》s16 = 8 * 17/2

》s16 = 68

************@@@@@**************

Answer 2

Answer:

The answer will be 68.

The above is the correct answer.☺