Question

What is the average of all numbers in the sequence 11, 13, 15,.........99 ?

Answer 1

Answer:

55.

Step-by-step explanation:

• 1 + 3 + ...........+ 9 =  n² [n = no. of odd no.]

          = 5² = 25

• 1 + 3 +.............+ 99 = 50² = 2500

So   11 + 13 + ..................+ 99 = 2500 - 25 = 2475

So average = 2475/45 = 55.

thank u.

Answer 2

Arithmetic progression is a progression in which every term after the first is obtained by adding a constant value, called the common difference (d). So, to find the nth term of an arithmetic progression, we know $an=a1+(n-1)d.$

Given series:

$11,13,15,..,99$

First term, $a_1=11$

Difference, $d=13-11=2$

Last term, $a_n=99$

Use formula

$a_n=a+(n-1)d$

$99=11+(n-1)2\\2(n-1)=99-11=88\\n-1=44\\n=45$

Sum of $45$ terms

$S_4_5=\frac{n}{2}[2a+(n-1)d]$

$=\frac{45}{2}[2\times 11+(45-1)2]$

$=\frac{45}{2}(22+44\times 2)\\ =45(11+44)\\=45\times 55\\S_4_5=2475$