Question

Find the ap whose third term is 16 and 7th term exceeds the 5th term by 12

Answer 1

Answer:

4, 10, 16, 22 ............

Step-by-step explanation:

a3 = a + 2d = 16. equ (1)

a7 = a + 6d

a5 = a + 4d

a7 - a5 = 12

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

a + 6d - a - 4d = 12

2d = 12

d = 12/2 = 6

substitute d in equ (1)

a + 2d = 16

a + 2(6) = 16

a + 12 = 16

a = 16 - 12

a = 4

the ap

a, a+d, a+2d, a+3d ........

4, 4+6, 4+12, 4+18 ......

4, 10, 16, 22 ........

Answer 2

$\small\tt\purple{Let\:the\:first\:term\:of\:the\:AP=a}$

$\small\sf\purple{Common\:difference=d}$

$\small\tt\purple{Third\:term=16}$

$\large\sf\red{a3=16\longrightarrow a+2d=16......(1)}$

$\small\tt\purple{7th\:term\:exceeds\:the\:5th\:term\:by\:12}$

$\therefore$$\large\tt\pink{a7=a5+12}$

$\longrightarrow$$\large\tt\pink{a+6d=a+4d+12}$

$\longrightarrow$$\large\tt\pink{2d=12}$

$\longrightarrow$$\large\tt\pink{d=6}$

$\large\sf\green{Putting\:the\:value\:of\:d\:in\:eq.(1),we\:get}$

$\large\tt\pink{a+2(6)=16}$

$\longrightarrow$$\large\tt\pink{a=4}$

$\small\sf\blue{Hence,the\:AP=a,a+d,a+2d,....=4,10,16.}$