Question

determine the AP whose third term is 16 and 7th term exceeds 5th term by12

Answer 1

Question : Determine the AP whose third term is 16 and 7th term exceeds 5th term by 12.

Answer : AP = 4, 10, 16

Step by step explanation :

Let a be the first term, a3 be the third term , a5 be the fifth term and a7 be the seventh term.

a3 = 16

a7 = a5 + 12..(1)

Let d be the common difference

So ,

a7 = a5 + d + d

a7 = 2d...(2)

From (1) and (2) ,

a5 + 12 = a5 + 2d

2d = 12

d = 6

Given that,

a3 = 16

a3 = a + 2d = 16

a + 12 = 16

a = 4

First term = 4

Second term = a + 4 = 4 + 6 = 10

Third term = 6 + 10 = 16

Hence, The AP is 4, 10, 16..

Answer 2

$\mathfrak{\huge{Answer:}}$

$\sf{We're\:Given\:That:}$

3rd term of an AP = $\sf{a_{3}}$ = 16

7th term of the AP = $\sf{a_{7} = a_{5} + 12}$

Now, let :-

a = first term

d = common difference

$\sf{Method:}$

We're not given the value of the 7th term. To make our calculations and our answer simple, we need to link everything ( every equation ) with $\sf{a_{7}}$.

$\sf{a_{7}}$ = $\sf{a_{5} + 12}$ ....(1)

Also, we know that 5 + 2 = 7, same ideology can be used here. The common difference = d, thus:-

$\sf{a_{7}}$ = $\sf{a_{5} + 2d}$ ...(2)

Keep (1) and (2) equal to each other, and then solve.

$\sf{a_{5} + 12}$ = $\sf{a_{5} + 2d}$

We know that :-

$\sf{a_{5} = a + 4d}$, put the value in the formula and solve :-

$\sf{a + 4d + 12 = a + 4d + 2d}$

=》 $\sf{12 = 2d}$

=》 d = 6

Put the value of a, in any of the formula and solve. I'll use :-

3rd term = $\sf{a_{3} = a + 2d}$ = 16

=》 $\sf{a + 2d = 16}$

=》 $\sf{a = 16 - 12}$

=》 a = 4

AP becomes :- 4, 4 + 6, 4 + 6 + 6, 4 + 6 + 6 + 6.....

$\tt{\huge{AP = 4, 10, 16, 22...}}$