Question

8. Determine the A.P. whose 3rd term is 16 and
the 7th term exceeds the 5th term by 12.​

Answer 1

EXPLANATION.

• GIVEN

3rd term of Ap = 16

a + 2d = 16 ....(1)

7th term exceeds the 5th term by 12

A7 = A5 + 12

A7 - A5 = 12

a + 6d - ( a + 4d ) = 12

a + 6d - a - 4d = 12

2d = 12

d = 6

put the value of d in equation (1)

we get,

a + 2(6) = 16

a + 12 = 16

a = 4

Nth term of Ap = An = a + ( n - 1 ) d

put the value of a and d in equation

4 + ( n - 1 ) 6

4 + 6n - 6

6n - 2

so, we can put the value of n in equation

put n = 1

6(1) - 2 = 4

put n = 2

6(2) - 2 = 10

put n = 3

6(3) - 2 = 16

put n = 4

6(4) - 2 = 22

Therefore, Ap can be written as,

4,10,16,22.......

Answer 2

Given:

t3 of an A.P. = 16 and

t7 = t5 + 12

Therefore:

tn = a + (n – 1)d

Thus:

t3 = a + (3 – 1)d = 16

a + 2d = 16 ________ (i)

And:

a + (7 – 1)d = a + (5 – 1)d + 12

6d = 4d + 12

2d = 12

d = 6

Using ‘d’ in (i) we get:

a + 2(6) = 16

a + 12 = 16

a = 4