show that p-7, p-10, p-13, p-16, p-19,....are in AP . find its 8th term and the common difference?
Answers
Answer: 8th Term = ( p - 28 ) .
Common difference = ( -3 ) .
Step-by-step explanation:
Consider :
p-7 as the 1st Term (T1) ,
p-10 as the 2nd Term (T2) ,
p-13 as the 3rd Term (T3) ,
p-16 as the 4th Term (T4) , and
p-19 as the 5th Term (T5) .
A is the variable which demonstrates the 1st Term (T1) . and ,
D is variable that gives us the uniform difference between two terms .
So ,
D = T2 - T1
Therefore , D = ( p-7 )-( p-10 )
D = - 3 .
D = T3 - T2
Therefore , D = ( p-13 )-( p-10 )
D = - 3 .
D = T4 - T3
Therefore , D = ( p-16 )-( p-13 )
D = - 3 .
D = T5 - T4
Therefore , D = ( p-19 )-( p-16 )
D = - 3 .
In each case , The value of D is uniform .
So , we can say that the terms are in uniform intervals , so they are in AP.
Term 8 (T8) = A + ( 7 * D )
Term 8 (T8) = p -7 + ( 7 * - 3 )
Term 8 (T8) = p -7 - 21
Term 8 (T8) = p - 28 .
Answer:Step-by-step explanation:
Answer: 8th Term = ( p - 28 ) .
Common difference = ( -3 ) .
Step-by-step explanation:
Consider :
p-7 as the 1st Term (T1) ,
p-10 as the 2nd Term (T2) ,
p-13 as the 3rd Term (T3) ,
p-16 as the 4th Term (T4) , and
p-19 as the 5th Term (T5) .
A is the variable which demonstrates the 1st Term (T1) . and ,
D is variable that gives us the uniform difference between two terms .
So ,
D = T2 - T1
Therefore , D = ( p-7 )-( p-10 )
D = - 3 .
D = T3 - T2
Therefore , D = ( p-13 )-( p-10 )
D = - 3 .
D = T4 - T3
Therefore , D = ( p-16 )-( p-13 )
D = - 3 .
D = T5 - T4
Therefore , D = ( p-19 )-( p-16 )
D = - 3 .
In each case , The value of D is uniform .
So , we can say that the terms are in uniform intervals , so they are in AP.
Term 8 (T8) = A + ( 7 * D )
Term 8 (T8) = p -7 + ( 7 * - 3 )
Term 8 (T8) = p -7 - 21
Term 8 (T8) = p - 28 .