Find the sum of 11 terms im an AP whose middle most term is 30
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Answered by
1
Answer:
Hope it helps
Step-by-step explanation:
Given middle term or the sixth term is 30
As we know, a n =a+(n−1)d, where a & d are the first term and common difference of an AP respectively.
Then, a+5d=30
Sum = n/2 (2a+(n−1)d)
= 11 /2 (2a+10d)
Substituting,
= 11/2 (2×30)
=330
Answered by
1
No of terms = 11 as it is even
Middle term will be(n + 1/2) i.e. 6th term
Let the first term be a and common difference be d
As an = a + (n - 1)d
a6 = a + (6 - 1)d
30 = a + 5d _____1
Now as we know
Sn = n/2 [2a+ (n-1) d]
S11 = 11/2 [2a + 10d]
= 11/2 [ 2(a + 5d)]
= 11/2 [2(30)] {from 1}
= 330
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