Question

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

Answer 1

$a_3$ = 16

$a_5 - a_7$ =12

$a_n$= a+(n-1)×d

=> $a_3$= a+(3-1)×d=16

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

similarly,

$a_7 \: and \: a_5$

will be a+6d and a+4d respectively...

then,

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

=>a+6d-a-4d=12

=>2d=12

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

=>d=6

putting d=6 in eq. (1)

=> a+2(6)=16

=> a+12=16

=> a=16-12

=> a=4

so, the A.P. will be,

=> 4,10,16,22...

$<marquee>Hope it helps!❤</marquee>︎︎$

Answer 2

$\underline{\sf{Answer\ :-}}$

$\sf{T3 = 16}$

$\implies{\sf{a + 2d = 16 -----(1)}}$

$\sf{T7 = T5 + 12}$

$\implies{\sf{a + 6d = a + 4d + 12}}$

$\implies{\sf{6d = 4d + 12}}$

$\implies{\sf{6d - 4d = 12}}$

$\implies{\sf{2d = 12}}$

$\implies{\sf{d = \dfrac{12}{2}}}$

$\implies{\sf{d = 6}}$

Substitute the value of d in (1) :-

$\implies{\sf{a + 2d = 16}}$

$\implies{\sf{a + 2 (6) = 16}}$

$\implies{\sf{a + 12 = 16}}$

$\implies{\sf{a = 16 - 12}}$

$\implies{\sf{a = 4}}$

So, a = 4.

$\sf{T1 = 4}$

$\sf{T2 = a + d}$

$\implies{\sf{4 + 6}}$

$\implies{\sf{10}}$

$\sf{T3 = a + 2d}$

$\implies{\sf{4 + 2 (6)}}$

$\implies{\sf{16}}$

Similarly, we can find all the terms of A.P.

So, the A.P. is :-

$\boxed{\sf{4, 10, 16, 22 ........}}$