Question

find the sum of the series 45 + 47 + 49 + dotdotdot + 99​

Answer 1

Step-by-step explanation:

a=45 d= 47-45=2

Tn=99

Tn=45+(n-1)2

99=45+2n-2

99-45+2=2n

56=2n

n=28

now

Sn=n/2(a+Tn)

Sn=28/2(45+99)

Sn=14(144)

Sn=2016

Hope it will help u out

Answer 2

Answer:

The first term in the Arithmetic Progression is 45 and the Common Difference for each term is 2, i.e. 45 + 2 = 47, 47 + 2 = 49 etc.

99 = a + (n - 1)d

44 = (n - 1) x 2

22 = n - 1

n = 23

(Here, l is the last term, a is the first term, and n is the number of terms in the series.)

Now, the last term is 99.

We know that sum of terms in an A.P. = n$\frac{a + l}{2}$

Here, l is the last term, a is the first term, and n is the number of terms in the series.

Sum of terms = $\frac{23(45 + 99)}{2}$

= $\frac{144 * 23}{2}$

= $\frac{72 * 23}{1}$

= 72 x 23

= 1656