Question

Question :-

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

Answer 1

Solution:

Let a be the 1st term and d be the common diffrence of A.P

We have,

$\bullet\:\sf a_3 = 16\\\\\\\bullet\:\sf a + 2d = 16\:\:\:\:\Bigg\lgroup\bf{Equation (1)}\Bigg\rgroup\\\\\\\bullet\:\sf a_7 = a + 6d\:and\:a_5 = a + 4d$

$\rule{130}1$

☯$\underline{\boldsymbol{According\: to \:the\: Question\:now :}}$

$:\implies\sf a_7 - a_5 = 12 \\\\\\:\implies\sf a + 6d - (a + 4d) = 12\\\\\\:\implies\sf a + 6d - a - 4d = 12\\\\\\:\implies\sf 2d = 12 \\\\\\:\implies\sf d = \dfrac{12}{2}\\\\\\:\implies\underline{\boxed{\sf d = 6}}$

$\rule{170}2$

$\underline{\bigstar\:\textbf{Putting the value of d in equation (1), we get :}}$

$\dashrightarrow\sf\:\:a + 2\times 6 = 16\\\\\\\dashrightarrow\sf\:\:a + 12 = 16\\\\\\\dashrightarrow\sf\:\:a = 16 - 12 \\\\\\\dashrightarrow\:\:\underline{\boxed{\sf a = 4}}$

So the first term is 4

We can find AP by adding d continuously.

$\therefore\:\underline{\textsf{Required AP is \textbf{4, 6, 10, 16......}}}.$

Answer 2

Am-An=(m-n)d
So 2d= 12
d= 6
A(3)=a+2d
Therefore a+12=16
a=4

APIs

4,10,16,22,——