Question

Find the arithmetic progression whose third term is 16 and seventh term exceeds its fifth term by 12.

Answer 1

Answer:

Step-by-step explanation:

AP is 4, 10, 16, 22, 28.......  

Step-by-step explanation:

Given :  

a3 = 16 , a7 = a5 + 12

Let 'a' be the first term and common difference be 'd'.

an = a + (n -1)d

a3 = a + (3 - 1)d

16 =  a + 2d ………..(1)

 

a7 = a5 + 12  

a + (7 - 1) d = a + (5 - 1) d + 12

a + 6d = a + 4d + 12

a  - a + 6d - 4d = 12

2d = 12

d = 12/2

d = 6  

On putting the value of d = 6 in equation 1

16 =  a + 2d

16 = a + 2 × 6

16 = a + 12

a = 16 - 12

a = 4  

First term ,a =  4  

Second term , a2 = (a + d) =  4 + 6 = 10

Third term , a3 = 16 (Given)

Fourth term , a4 = (a + 3d) = 4 + 3×6

a4 = 4 + 18 = 22

Fifth term ,a5 = a + 4d = 4 + 4 × 6 = 4 + 24 = 28

Hence , AP is 4, 10, 16, 22, 28.......  

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Answer 2

$\huge{\underline{\underline{\bf{\red{Answer}}}}}$


AP is 4, 10, 16, 22, 28.......  

Step-by-step explanation:

Given :  

a3 = 16 , a7 = a5 + 12

Let 'a' be the first term and common difference be 'd'.

an = a + (n -1)d

a3 = a + (3 - 1)d

16 =  a + 2d ………..(1)

 

a7 = a5 + 12  

a + (7 - 1) d = a + (5 - 1) d + 12

a + 6d = a + 4d + 12

a  - a + 6d - 4d = 12

2d = 12

d = 12/2

d = 6  

On putting the value of d = 6 in equation 1

16 =  a + 2d

16 = a + 2 × 6

16 = a + 12

a = 16 - 12

a = 4  

First term ,a =  4  

Second term , a2 = (a + d) =  4 + 6 = 10

Third term , a3 = 16 (Given)

Fourth term , a4 = (a + 3d) = 4 + 3×6

a4 = 4 + 18 = 22

Fifth term ,a5 = a + 4d = 4 + 4 × 6 = 4 + 24 = 28

Hence , AP is 4, 10, 16, 22, 28.......  



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