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

1.	Determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12.
Answer» $\bf{\underline{Solution}}:-$ Let first TERM of AP be a and COMMON difference be d Then, $\rm\:a_3=<klux>16</klux>$ ⤇ a + 2d = 16 ⤇ a = 16 - 2d ..............................(i) According to the question, $\rm\:a_7=a+4d+12$ ⤇ a + 6d = a + 4d + 12 ⤇ 6d = 4d + 12 ⤇ 6d - 4d = 12 ⤇ 2d = 12 ⤇ d = $\rm\cancel\dfrac{12}{2}$ ⤇ d = 6 putting the VALUE of d in eq(i) a = 16 - 2d ⤇ a = 16 - 2 × 6 ⤇ a = 16 - 12 ⤇ a = 4 Now, $\rm\:a_1=4$ $\rm\:a_2=a+d$ $\rm\longrightarrow\:a_2=4+6$ $\rm\longrightarrow\:a_2=10$ $\rm\:a_3=a+2d$ $\rm\longrightarrow\:a_3=4+2\times{6}$ $\rm\longrightarrow\:a_3=4+12$ $\rm\longrightarrow\:a_3=16$ $\rm\:a_4=a+3d$ $\rm\longrightarrow\:a_4=4+3\times{6}$ $\rm\longrightarrow\:a_4=4+18$ $\rm\longrightarrow\:a_4=22$ Hence,AP will be 4,10,16,22,......