Question

Determine an AP where 3d term is 16 and when 5term is subtracted from 7 term, we get 12.​

Answer 1

$\large\underline\mathtt{Answer:}$

Let the first term be, $a$ & Common Difference,$d$

3rd term = a + 2d =16

5th term = a + 4d

7th term = a + 6d

7th term - 5th term = 12

a + 6d - a + 4 d = 12

2 d = 12

d = $\frac{12}{2}$= 6

a + 2d = 16

a + 2 × 6 = 16

a = 4

Hence the AP is 4, 10, 16, 22, 28, 34, 40......

Answer 2

Answer:

ARITHMETIC PROGRESSIONS :

3rd term,

A3 = 16

16 = a + 2d --> ( i )

A7 - A5 = 12

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

a + 6d - a - 4d = 12

2d = 12

d = 12/ 2 = 6

d = 6

Now, putting value of 'd' in equation ( i ),

a + 2*6 = 16

a = 16 - 12

a = 4

Now, A. P. = a, a + d, a + 2d.....

A. P. = 4, 10, 16, 22.....