Question

Find the sum of the first 12 terms of the arithmetic sequence. 50,47,44,41,38.....
Answer this with solution pleasee thank you in advance

Answer 1

Sn=n/2× (2a+(n-1)d)
Sn=12/2×(2×50+(12-1)-3)
=6×(100-33)
=6×67
=402.

Answer 2

Given:

• An arithmetic series = 50, 47, 44, 41, 38,...

To Find:

• The sum of the first 12 terms of the given arithmetic series.

Solution:

The formula to find the sum of "n" terms of an arithmetic series is,

⇒ $S_n$ = n/2[2a+(n-1)d] → {equation 1}

Where "n" is the number of terms, "a" is the first term of an arithmetic series, and "d" is the difference between the terms.

From the given arithmetic series we get a few values. They are:

"n=12", "a=50", "d=-3"

Substitute the obtained values in equation 1.

The sum of the first 12 terms of an arithmetic series is given by,

⇒ $S_{12}$ = 12/2[(2×50)+(12-1)×(-3)]

In the above equation we solved for the terms in the bracket.

⇒ $S_{12}$ = 6[100-33]  {subtracting the terms}

⇒ $S_{12}$  = 6[67]  = 402 { multiplying the terms}

∴ The sum of the first 12 terms of an arithmetic series = 402