Find the sum of all 11 term of the ap whose middle term is 30
Answers
Answer:
330
Step-by-step explanation:
therefore the middle term is 6th term
a6= 30
a+5d= 30...(i)
sn= 11/2(2a+10d..ii
multiply equation (i) by 2
2a+10d= 60...(iii)
put value of iii in ii
now 11/2(60)
11 x 30
330 is the answer
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Question:
Find the sum of all 11 term of the ap whose middle term is 30.
Answer:
330
Note:
• A sequence is said to be in AP (Arithmetic Progression), if the difference between its consecutive terms are equal.
• The nth term of an AP is given as ;
T(n) = a + (n-1)•d , where a is the first term and d is the common difference.
• The common difference of an AP is given as ;
d = T(n) - T(n-1)
• If the number of terms in an AP is n ( where n is odd ) ,then there will be a single middle term.
Also, [(n+1)/2]th term will be its middle term.
• If the number of terms in an AP is n ( where n is even ) ,then there will be two middle terms.
Also, (n/2)th and (n/2 + 1)th terms will be its middle terms.
• The sum up to nth terms of an AP is given as ;
S(n) = (n/2)•[2a + (n-1)•d] where a is the first term and d is the common difference.
• The nth term of an AP is also given as ;
T(n) = S(n) - S(n-1)
Solution:
Given:
Total number of teams in the AP is 11.
( ie; n = 11 )
Middle term of the AP is 30.
To find:
The sum of all the 11 terms of the AP.
( ie; S(11) = ? )
Here,
n = 11 (which is an odd number)
Thus,
=> Middle term = [(n+1)/2]th term
=> Middle term = [(11+1)/2]th term
=> Middle term = [12/2]th term
=> Middle term = 6th term
=> Middle term = T(6)
=> 30 = a + (6-1)•d {given; middle term = 30}
=> 30 = a + 5d
=> a + 5d = 30 ----------(1)
Now,
The sum of 11 terms of the AP will be given as;
=> S(11) = (11/2)•[2a + (11-1)•d]
=> S(11) = (11/2)•[2a + 10d]
=> S(11) = (11/2)•2•[a + 10d]
=> S(11) = (11•2/2)•[a + 10d]
=> S(11) = 11•[a + 10d]
=> S(11) = 11•30 {using eq-(1)}
=> S(11) = 330
Hence,
The sum of all the 11 terms of the AP is 330.