The first term of an a.p is 2 and the common difference is 3 find the sum of the first 11 terms
Sn= n/2[2a+(n-1)d]
S11= 11/2[4+(11-1)3]
=11/2(4+30)
=11/2(34)
=187
11th term is ... 2 + (10 * 3) = 32
sum of 1st eleven terms ... [(2 + 32) / 2] * 11
To find the sum of an arithmetic progression (AP), you can use the formula:
Sum = (n/2) * [2a + (n-1)d]
where:
- n is the number of terms,
- a is the first term, and
- d is the common difference.
In this case, the first term (a) is 2, the common difference (d) is 3, and we need to find the sum of the first 11 terms (n = 11).
Substituting the given values into the formula, we have:
Sum = (11/2) * [2(2) + (11-1)(3)]
= (11/2) * [4 + 10(3)]
= (11/2) * [4 + 30]
= (11/2) * 34
= 11 * 17
= 187
Therefore, the sum of the first 11 terms of the arithmetic progression is 187.
To find the sum of the first 11 terms of an arithmetic progression (AP), you can use the formula:
Sum of AP = (n/2) * [2a + (n - 1)d]
Where:
- n is the number of terms
- a is the first term
- d is the common difference
In this case, the first term (a) is 2, the common difference (d) is 3, and n is 11.
Substituting these values into the formula, we get:
Sum of AP = (11/2) * [2(2) + (11 - 1)(3)]
Simplifying further:
Sum of AP = (11/2) * [4 + 10(3)]
Sum of AP = (11/2) * [4 + 30]
Sum of AP = (11/2) * [34]
Sum of AP = 11 * 17
Sum of AP = 187
Therefore, the sum of the first 11 terms of the arithmetic progression is 187.