Question

The sequence of 28,34,40,46,52,...,88 has 11 terms. find the sum of the 11 terms

Answer 1

Answer:

638

Step-by-step explanation:

The sum of the first 11th term of the sequence is 638

The sequence in question is an arithmetic progression. The formula for finding the sum of an A.P is expressed as:

S_n=\frac{n}{2}(2a+(n-1)d)S

n

=

2

n

(2a+(n−1)d)

a is the first term = 28

d is the common difference = 34 - 28 = 40 - 34 = 6

n is the number of terms = 11

Substitute the given parameters into the expression

\begin{gathered}S_{11}=\frac{11}{2}(2(28)+(11-1)(6))\\S_{11}=\frac{11}{2}(56+(10((6))\\S_{11}=\frac{11}{2}(116)\\S_{11}=11\times 58\\S_{11}=638\end{gathered}

S

11

=

2

11

(2(28)+(11−1)(6))

S

11

=

2

11

(56+(10((6))

S

11

=

2

11

(116)

S

11

=11×58

S

11

=638

Hence the sum of the first 11th term of the sequence is 638

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Answer 2

Answer:

Sum of the 11 term is 638 .

Step-by-step explanation:

Explanation:

Given, 28,34,40,46,52.......88

which has 11 term.

We know , $S_{n} = \frac{n}{2} [2a+(n-1)d]$

Where , n = number of term

a = first term  of the given sequence

d = common difference  between two consecutive number .

Step 1:

Therefore we have , a = 28 , d= 6 and n = 11 now put all this value in the sum formula

$S_{11} = \frac{11}{2} [2(28)+(11-1)6]$

         =$\frac{11}{2}[56+60]$

           =$\frac{11}{2}[116] = 638$

Final answer :

Hence , the sum of the 11term is 638.