Answer :
The sum of 11 terms of the given A.P. is 660.
To find the sum of 11 terms of an arithmetic progression (A.P.), we first need to identify the first term (a) and the common difference (d). Since the middle term is given as 30, we can consider it as the sixth term of the A.P.
In an arithmetic progression with an odd number of terms, the middle term is the median, which can be calculated as:
[tex]\[ \text{Middle term} = \frac{\text{First term} + \text{Last term}}{2} \][/tex]
Given that the middle term is 30, we can write this as:
[tex]\[ 30 = \frac{a + (a + 10d)}{2} \][/tex]
Solving this equation, we find:
[tex]\[ a + 10d = 60 \]\[ a = 60 - 10d \][/tex]
Now, we can use the formula for the sum of an arithmetic series:
[tex]\[ S = \frac{n}{2} \times (a + l) \][/tex]
where S is the sum of the series, n is the number of terms, a is the first term, and I is the last term.
Since the A.P. has 11 terms, the last term is the 11th term. Therefore, l = a + 10d.
Substituting the expressions for a and l into the sum formula:
[tex]\[ S = \frac{11}{2} \times (a + (a + 10d)) \]\[ S = \frac{11}{2} \times (2a + 10d) \][/tex]
Substituting a = 60 - 10d into the equation:
[tex]\[ S = \frac{11}{2} \times (2(60 - 10d) + 10d) \]\[ S = \frac{11}{2} \times (120 - 10d + 10d) \]\[ S = \frac{11}{2} \times 120 \]\[ S = \frac{11 \times 120}{2} = \frac{1320}{2} = 660 \][/tex]
Therefore, the sum of 11 terms of the given A.P. is 660.
Question:-
The sum of 11 terms of an A.P. whose middle term is 30, is ______
